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Mathematics III
Unit 3
2nd Edition
Mathematics III
Frameworks
Teacher’s Edition
Unit 3
Exploring Exponentia
(Exponents and
Logarithms)
2nd Edition
June, 2010
Georgia Department of Education
Georgia Department of Education
Kathy Cox, State Superintendent of Schools
June, 2010
Copyright 2009 © All Rights Reserved
Unit 3: Page 1 of 66
Mathematics III
Unit 3
2nd Edition
Table of Contents
INTRODUCTION: ......................................................................Error! Bookmark not defined.3
The Planet of Exponentia Learning Task........................................................................................ 7
How Long Does it Take? Learning Task ...................................................................................... 15
The Population of Exponentia Learning Task .............................................................................. 26
Modeling Natural Phenomena on Earth Learning Task ............................................................... 43
Culminating Task: Traveling to Exponentia ................................................................................. 59
Georgia Department of Education
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June, 2010
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Unit 3: Page 2 of 66
Mathematics III
Unit 3
2nd Edition
Mathematics III – Unit 3
Exploring Exponentia
(Exponents and Logarithms)
Teacher’s Edition
INTRODUCTION:
In Math 2, students learned about exponential functions, including the natural exponential
function, e. Additionally, throughout the previous grades and units, students have explored basic
functions, inverse functions, and the impact of different transformations on the graphs of
functions. This unit builds on students’ understandings of inverse functions, exponential
functions, and transformations of functions. Before delving into logarithmic functions, it is
important for students to understand nth roots and rational exponents, so the study of these topics
is motivated in the first tasks. Review of the essential prerequisites is used as a way to motivate
the study of logarithms, primarily common and natural logarithms. This unit is not concerned
with the properties of logarithms or with solving complex logarithmic equations. Rather, it
focuses on building student conceptual understandings of logarithms.
ENDURING UNDERSTANDINGS:






Nth roots are inverses of power functions. Understanding the properties of power
functions and how inverses behave explains the properties of nth roots.
Real-life situations are rarely modeled accurately using discrete data. It is often necessary
to introduce rational exponents to model and make sense of a situation.
Computing with rational exponents is no different from computing with integral
exponents.
Logarithmic functions are inverses of exponential functions. Understanding the properties
of exponential functions and how inverses behave explains the properties and graphs of
logarithms.
Exponential and logarithmic functions behave the same as other functions with respect to
graphical transformations.
Two special logarithmic functions are common logarithms and natural logarithms. These
special functions occur often in nature.
KEY STANDARDS ADDRESSED:
MM3A2. Students will explore logarithmic functions as inverses of exponential functions.
a. Define and understand the properties of nth roots.
b. Extend the properties of exponents to include rational exponents.
c. Define logarithmic functions as inverses of exponential functions.
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Unit 3: Page 3 of 66
Mathematics III
Unit 3
2nd Edition
e. Investigate and explain characteristics of exponential and logarithmic functions including
domain and range, asymptotes, zeros, intercepts, intervals of increase and decrease, and
rate of change.
f. Graph functions as transformations of f(x) = ax, f(x) = logax, f(x) = ex, and f(x) = ln x.
g. Explore real phenomena related to exponential and logarithmic functions including halftime and doubling time.
RELATED STANDARDS ADDRESSED:
MM3A1. Students will analyze graphs of polynomial functions of higher degree.
d. Investigate and explain characteristics of polynomial functions, including domain and
range, intercepts, zeros, relative and absolute extrema, intervals of increase and decrease,
and end behavior.
MM3A3. Students will solve a variety of equations and inequalities.
b. Solve polynomial, exponential, and logarithmic equations analytically, graphically, and
using appropriate technology.
d. Solve a variety of types of equations by appropriate means choosing among mental
calculation, pencil and paper, or appropriate technology.
MM1P1. Students will solve problems (using appropriate technology).
a. Build new mathematical knowledge through problem solving.
b. Solve problems that arise in mathematics and in other contexts.
c. Apply and adapt a variety of appropriate strategies to solve problems.
d. Monitor and reflect on the process of mathematical problem solving.
MM1P2. Students will reason and evaluate mathematical arguments.
a. Recognize reasoning and proof as fundamental aspects of mathematics.
b. Make and investigate mathematical conjectures.
c. Develop and evaluate mathematical arguments and proofs.
d. Select and use various types of reasoning and methods of proof.
MM1P3. Students will communicate mathematically.
a. Organize and consolidate their mathematical thinking through communication.
b. Communicate their mathematical thinking coherently and clearly to peers, teachers, and
others.
c. Analyze and evaluate the mathematical thinking and strategies of others.
d. Use the language of mathematics to express mathematical ideas precisely.
MM1P4. Students will make connections among mathematical ideas and to other
disciplines.
a. Recognize and use connections among mathematical ideas.
b. Understand how mathematical ideas interconnect and build on one another to produce a
coherent whole.
c. Recognize and apply mathematics in contexts outside of mathematics.
Georgia Department of Education
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June, 2010
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Unit 3: Page 4 of 66
Mathematics III
Unit 3
2nd Edition
MM1P5. Students will represent mathematics in multiple ways.
a. Create and use representations to organize, record, and communicate mathematical ideas.
b. Select, apply, and translate among mathematical representations to solve problems.
c. Use representations to model and interpret physical, social, and mathematical
phenomena.
Unit Overview:
The launching task reviews a number of ideas students learned in Math 2 that are essential
understandings for this unit. Specifically, knowledge and facility with the content in MM2A5 is
necessary for completing this task. Students are required to recall knowledge about the
relationships between functions and their inverses. Students must also recall the surface area and
volume of spheres, another topic from Math 2. Beginning in this task and continuing throughout
the unit is an emphasis on multiple representations, i.e. using tables, graphs, and equations.
The second task continues reviewing content, primarily exponential functions, and motivates
the need for determining the inverse of an exponential function. Also addressed is the need for
rational exponents in working with exponential equations. The properties of exponents are
reviewed and students practice working with rational exponents. Solving exponential equations
graphically is stressed in this task. Students also determine half-lifes and doubling time in the
task.
Task 3 builds on students’ understandings of exponential functions and inverses to develop
the idea of a logarithm. Through paper folding and examining graphs, students determine the
graph of the inverse of an exponential function, eventually assigning the inverse the name
logarithm. Students then work with logarithms and their graphs, including common logarithms,
to deepen their understanding and solve exponential problems.
In the fourth task, students model natural phenomena with exponential and logarithmic
functions, including natural logarithms. The concept of natural logarithm is developed in context.
Throughout the task, students are asked to consider how the equations of natural phenomena are
transformations of the basic exponential or logarithm functions.
Finally, the culminating task requires the students to use their knowledge from the unit to
answer a variety of questions about naturally occurring phenomena. Many of the questions
resemble those asked in the learning tasks, but they generally possess a slightly higher level of
complexity. To close the unit, students are asked to use graphs of the functions in this unit to
design an alien creature encountered during the unit.
Throughout this unit, including the culminating task, it is assumed that students have
access to a graphing utility. Tasks stress knowing the domain and range of functions, partially as
a way to help students learn to set their graphing window and partially as a way to judge the
validity of their answers. The unit emphasizes the reasonableness of results and models. Students
should be encouraged to communicate their reasoning both in writing and in class discussions.
Georgia Department of Education
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June, 2010
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Mathematics III
Unit 3
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Vocabulary and formulas:
Asymptote: A line or curve that describes the end behavior of the graph. A graph never crosses a
vertical asymptote but it may cross a horizontal or oblique asymptote.
Common logarithm: A logarithm with a base of 10. A common logarithm is the power, a, such
that 10a = b. The common logarithm of x is written log x. For example, log 100 = 2 because 102
= 100.
Compounded continuously: Interest that is, theoretically, computed and added to the balance of
an account each instant. The formula is A = Pert, where A is the ending amount, P is the principal
or initial amount, r is the annual interest rate, and t is the time in years.
Compounded interest: A method of computing the interest, after a specified time, and adding
the interest to the balance of the account. Interest can be computed as little as once a year to as
many times as one would like. The formula is A = P0(1 + r/n)nt, where A is the ending amount,
Po is the initial amount, r is the annual interest rate, n is the number of times compounded per
year, and t is the number of years.
Exponential functions: A function of the form y = a·bx where a > 0 and either 0 < b < 1 or b >
1.
Logarithmic functions: A function of the form y = logbx, with b  1 and b and x both positive.
A logarithmic function is the inverse of an exponential function. The inverse of y = bx is y =
logbx.
Logarithm: The logarithm base b of a number x, logbx, is the power to which b must be raised to
equal x.
Natural exponential: Exponential expressions or functions with a base of e, i.e. y = ex.
Natural logarithm: A logarithm with a base of e. A natural logarithm is the power, a, such that
ea = b. The natural logarithm of x is written ln x. For example, ln 8 = 2.0794415… because
e2.0794415…= 8.
Nth roots: The number that must be multiplied by itself n times to equal a given value. The nth
root can be notated with radicals and indices or with rational exponents, i.e. x1/3 means the cube
root of x.
m
n
Rational exponents: For a > 0, and integers m and n, with n > 0, a n  a m 
 a ; a
n
m
m/n
=
(a1/n)m = (am)1/n .
Transcendental numbers: Real numbers that are not algebraic. That is, they are not solutions to

polynomial equations with integer coefficients. Two examples
are e and π.
Georgia Department of Education
Kathy Cox, State Superintendent of Schools
June, 2010
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Unit 3: Page 6 of 66
Mathematics III
Unit 3
2nd Edition
Notes on The Planet of Exponentia Learning Task
This task launches the unit by reviewing students’ understanding of function and inverses of
functions. The relationships between functions and their inverses; in terms of domain and
range, graphs, and equations; will be reviewed. Towards the end, the task begins building the
idea of rational exponents, as a new way for representing roots of expressions when
determining the inverse of functions. The review was included because it is extremely
important for students to understand inverses to understand the way logarithmic functions are
developed in the subsequent tasks.
This task reviews standards MM2A5 and introduces standards MM3A2a and MM3A2b.
Supplies Needed:
 Graph paper
 Calculator or computer program such as Excel
LAUNCHING TASK: THE PLANET OF EXPONENTIA
A new solar system was discovered far from the Milky Way in 1999. One of the planets in the
system, Exponentia, has a number of unique characteristics. Scientists noticed that the radius of
the planet has been increasing 500 meters each year.
When NASA scientists first spotted Exponentia, its diameter was approximately 40 km.
1. Make a table that lists the diameter and surface area of the planet from the years 1999 to
2009. (Leave your surface area answers in terms of π.)
1999
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
Diameter (in km)
40
Surface Area (in km2)
Solution: (This is an ideal place to use graphing calculators and the lists functions.)
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June, 2010
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Mathematics III
Unit 3
Diameter (in km)
40
41
42
43
44
45
46
47
48
49
50
2nd Edition
Surface Area (in km2)
1600 π
1681 π
1764 π
1849 π
1936 π
2025 π
2116 π
2209 π
2304 π
2401 π
2500 π
a. Write a function rule that expresses the relationship between the radius of the
planet and its surface area. What does it mean for the surface area to be a function
of the radius (or diameter)? (Make sure you use proper function notation.)
Solution: SA = 4πr2  f(r)= 4πr2; If the surface area is a function of the radius then we know
that (1) the equation for surface area of a sphere is a function; for each radius, only one
surface area can be determined; and (2) that surface area can be determined by knowing the
radius. That is, the surface area is the dependent variable and radius in the independent
variable.
b. Interchange the columns and create a second table so that surface area is the
independent variable and diameter is the dependent.
Surface Area (in km2)
1999
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
Diameter (in km)
40
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Unit 3: Page 8 of 66
Mathematics III
Unit 3
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Solution:
Surface Area (in km2)
1600 π
1681 π
1764 π
1849 π
1936 π
2025 π
2116 π
2209 π
2304 π
2401 π
2500 π
Diameter (in km)
40
41
42
43
44
45
46
47
48
49
50
c. Graph the data from the first table. (Unless you are graphing on a calculator or
computer, graphing every other point is sufficient.) How would a graph of the
data from the second table look? How do you know?
Solution:
The horizontal and vertical axes would be switched in the second graph. This graph is slightly
upturned. The other one would be curved slightly down. Because we switched the independent
and dependent variables, the second is the graph of the inverse. The graphs of inverses are
reflections of each other over the line y = x.
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June, 2010
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Mathematics III
Unit 3
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d. If we wrote a rule (equation) for the new relationship in part (b), how would the
new rule be related to the original? That is, how are the two rules related to each
other? How do you know?
Solution: Because we switched the independent and dependent variables, the second is the
inverse of the first.
e. Using algebra, write a rule for the data in the second table. (Hint: We want an
equation for the radius in terms of the surface area.)
Solution: Algebraically determining the inverse, the answer is r  
SA
. Restrictions will be
4
addressed in the next part.
f. What are the domain and range of the function in part (a)? What are the domain
 What are the restrictions on the domain
and range for the new relation in part (e)?
and range due to the context of the problem? Why are there restrictions?
Function in Part (a): Domain ______________
Range ______________________
Restrictions due to context:
Relation in Part (e): Domain ______________
Range ______________________
Restrictions due to context:
Solution:
Function in Part (a): Domain _All real numbers __
Range __All non-negative real numbers {SA: SA ≥ 0}
Restrictions due to context: The radius of the planet could never be negative. So,
technically the domain is {r: r ≥ 0} .[One might also say that since the smallest radius in our
data set is 20 km, that the domain is {r: r ≥ 20 km}.]
Relation in Part (e): Domain _ All non-negative real numbers {SA: SA ≥ 0} ___
Range __ All real numbers ________
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June, 2010
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Mathematics III
Unit 3
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Restrictions due to context: The radius of the planet could never be negative. So,
technically the range is {r: r ≥ 0}. In this case, we could change the relation in part (e) to
SA
r
. [One might also say that since the smallest radius in our data set is 20 km, that the
4
range is {r: r ≥ 20 km}.]

g. Is the new relation in part (e) also a function? How do you know? Explain two
ways: using the graph of the original function in part (a) and using the graph of
the unrestricted relation in part (e). (You may need to graph your equations on
your graphing calculator or computer.)
Solution: The new relation, as written with the plus/minus is not a function. If we restrict the
domain, as we would need to do due to the context, it would be a function. Looking at a graph
of the original surface area equation, the inverse – relation in part (e) – is not a function. The
original graph does not pass the horizontal line test. That is, for each output value (SA), there
are two input values (r). Graphing the relation in part (e) shows a sideways parabola. It does
not pass the vertical line test. That is, for each input (SA), there are two outputs (r).
2. Now let’s consider how the volume of the planet changes.
a. Write a function rule that expresses the relationship between the radius of the
planet and its volume.
4
4
Solution: V  r 3  f (r)  r 3
3
3

b. Graph the function from part (a) over the interval -10 ≤ r ≤ 10. What part, if any,
of this graph makes sense in the context of Exponentia’s volume? Explain.
Solution: The part of the graph that makes sense in
the context of the problem is when the radius is
greater than or equal to zero. The radius of the
planet cannot be negative.
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Mathematics III
Unit 3
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c. Using your exploration from part 1, consider the following.
i. If you wrote an equation with volume as the independent variable and the
radius as the dependent variable, would this be a function? Explain.
Solution: Yes, this new equation would represent a function. The volume passes the horizontal
line test, so the inverse would pass the vertical line test. Even in the new equation, there would
only be one output for each input value.
ii. Describe how the graph of the new equation would look. Sketch this new
graph with your graph in part (b). (Hint: How are the graphs of inverses
related to each other?)
Solution: The coordinates of the
graph would be switched. The graphs
of inverse functions are reflections
over the line y = x.
d. We want to write a rule for the second relation.
i. Explain how you knew to find the equation in part 1(e).
Solution: In inverses, the inputs and outputs are reversed. Solving the equation for the
independent variable created the inverse.
ii. We can use similar reasoning in for finding the equation in this problem.
Let’s start by solving for r3.
4
3
3V
Solution: V  r 3  V  r 3 
 r3
3
4
4

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Mathematics III
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iii. Now, to finish solving, we need the inverse of r3. In number 1, to solve for
r2 by taking the square root of both sides of the equation. Likewise, we
take the cube root of both sides of the equation in the above step. Solve for
r.
Solution: r  3
3V
.
4
When we take the square root of an expression or number, such as x2 = 4, we must consider both
the positive and negative roots of the expression or number, so x =  2. Do we need to consider
both positive and negative roots when we take cube roots? Why or why not?
Solution: We do not need to consider positive and negative roots when taking cube roots. If the
radicand is positive, the cube root is positive. If the radicand is negative, the cube root is
negative. The product of three positive/negative numbers is positive/negative.
3. nth Roots:
The cube root of b is the number whose cube is b. Likewise, the nth root of b is the number that
when raised to the nth power is b. For example, the 5th root of 32 is 2 because 25 = 32. We can
write the 5th root of 32 as 5 32 .
Another notation used to represent taking roots employs exponents. Instead of writing the 5th root
1
1
of 32 as 5 32 , we can write it as 32 5 . The cube root of 27 can be written as 27 3 . How do you
think we would represent the nth root of a number x?
Solution: x
1
n
1
For the remainder of (3), consider f(x) = xn and g x   x n .
a. How do you think the graphs of f(x) and g(x) are related? Using graph paper
and/or a calculator, test your conjecture, letting n = 1, 2, 3, 4.
Solution: (The graphs are not provided here.) The graphs of f(x) and g(x) are reflections of
each other over the line y = x. So f(x) and g(x) are inverses.
b. Evaluate f(g(x)) and g(f(x)). If you need to, use your examples of n = 1, 2, 3, 4 to
help you determine these compositions. What do the results tell you about f(x) and
g(x)?
Solution: (The examples are not provided here.) Both compositions yield x. So f(x) and g(x)
are inverses.
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Mathematics III
Unit 3
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c. Explain how your work in problems 1 and 2 confirms your conclusions in parts
(a) and (b) of this problem.
Solution: Answers will vary but should include mention of the square root and quadratic
being inverses and the cube root and cube being inverses.
4. Using your investigations above and what you remember from Math 2, write a paragraph
summarizing characteristics of inverses of functions, how to find inverses algebraically
and graphically, and how to tell if inverses are functions.
Solution: Answers will vary but should include that the domain and range of inverse functions
are interchanged; their graphs are reflections over the line y = x; they are found algebraically
(in functions written in terms of x and y) by interchanging the variables and solve for the new
dependent variable; horizontal and vertical line tests (and how they determine if the inverses
are functions – one-to-one functions have functional inverses); and that the composition of a
function and its inverse yields f(g(x)) = x.
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Mathematics III
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Notes on How Long Does It Take? Learning Task
Although this task uses the same context as the launching task, it is not necessary to have
completed the launching task to complete this task. Students are expected to recall knowledge
of writing exponential functions for this task. The task develops and provides instruction on
standards MM3A2a and MM3A2b. Standards MM3A2c, MM3A2d, and MM3A2g will also be
addressed although not fully developed. Specifically, the need for finding the inverse of
exponential expressions will be established (element c), students will review characteristics of
exponential functions (element e), and students will explore phenomena related to exponential
functions (element g).
Supplies Needed:
 Graphing Calculator
 Graph paper
HOW LONG DOES IT TAKE?
Before sending astronauts to investigate the new planet of Exponentia, NASA decided to run a
number of tests on the astronauts.
1. A specific multi-vitamin is eliminated from an adult male’s bloodstream at a rate of about
20% per hour. The vitamin reaches peak level in the bloodstream of 300 milligrams.
a. How much of the vitamin remains 2 hours after the peak level? 5 hours after the
peak level? Make a table of values to record your answers. Write expressions for
how you obtain your answers.
Time (hours) since peak
0
1
2
3
4
5
300
Vitamin concentration in
bloodstream (mg)
Solution:
Time (hours) since
peak
0
1
2
300
240
192
3
4
5
153.6 122.88 98.304
Vitamin concentration
in bloodstream (mg)
300*(1 -.2) = 240; 240 * .8 = 192; 192 * .8 = 153.6; 153.6 * .8 = 122.88; 122.88 * .8 = 98.304
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Mathematics III
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b. Using your work from (a), write expressions for each computed value using the
original numbers in the problem (300 and 20%). For example, after 2 hours, the
amount of vitamin left is [300 * (1 - .2)]* (1 - .2).
Solution:
After 1 hour: 300 * (1 - .2) = 240; after 2 hours: 300 * (1 - .2)2 = 192;
After 3 hours: 300 * (1 - .2)3 = 153.6; after 4 hours: 300 * (1 - .2)4 = 122.88;
After 5 hours: 300 * (1 - .2)5 = 98.304
c. Using part (b), write a function for the vitamin level with respect to the number of
hours after the peak level, x.
Solution: f(x) = 300(1 - .2)x or f(x) = 300(.8)x
d. How would use the function you wrote in (c) to answer the questions in (a)? Use
your function and check to see that you get the same answers as you did in part
(a).
Solution:
f(0) = 300(.8)0 = 300
f(1) = 300(.8)1 = 240
f(2) = 300(.8)2 = 192
f(3) = 300(.8)3 = 153.6
f(4) = 300(.8)4 = 122.88
f(5) = 300(.8)5 = 98.304
e. After how many hours will there be less than 10 mg of the vitamin remaining in
the bloodstream? Explain how you would determine this answer using both a
table feature and a graph.
Solution: Between 15 and 16 hours after peak, the concentration dips below 10 mg. Using the
table feature, at 15 hours, the concentration is 10.555 mg whereas at 16 hours, it is 8.4442.
Using a graph, we can trace the graph to determine where the vitamin concentration dips
below 10 mg. This occurs around 15.3 hours.
f. Write an equation that you could solve to determine when the vitamin
concentration is exactly 10 mg. Could you use the table feature to solve this
equation? Could you use the graph feature? How could you use the intersection
feature of your calculator? Solve the problem using one of these methods.
Solution: 300(.8)x = 10; You could use the table feature to approximate the solution by making
smaller and smaller table steps. The trace feature could also be used to approximate the
solution. If the original equation was graphed as Y1 and you let Y2 = 10, we could look for the
intersection. Using this last method, the solution is 15.242194 hours or approximately 15
hours, 14 minutes, 32 seconds.
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g. How would you solve the equation you wrote in (f) algebraically? What is the
first step?
Solution: The first step would be to divide by 300.  .8x = 1/30.
To finish solving the problem algebraically, we must know how to find inverses of exponential
functions. This topic will be explored later in this unit.
2. A can of Instant Energy, a 16-ounce energy drink, contains 80 mg of caffeine. Suppose
the caffeine in the bloodstream peaks at 80 mg. If ½ of the caffeine has been eliminated
from the bloodstream after 5 hours (the half-life of caffeine in the bloodstream is 5
hours), complete the following:
a. How much caffeine will remain in the bloodstream after 5 hours? 10 hours? 1
hour? 2 hours? Make a table to organize your answers. Explain how you came up
with your answers. (Make a conjecture. You can return to your answers later to
make any corrections.)
Time (hours) since
peak
0
1
2
5
10
5
10
40
20
80
Caffeine in
bloodstream (mg)
Solution:
Time (hours) since
peak
0
80
1
2
69.644 60.629
Caffeine in
bloodstream (mg)
(Explanations will vary. At this point, students may not have the answers for 1 hour and 2
hours correct. They will be able to go back and correct it.)
b. Unlike problem (1), in this problem in which 80% remained after each hour, in
this problem, 50% remains after each 5 hours.
i. In problem (1), what did the exponent in your equation represent?
Solution: The exponent represented the number of hours that had passed.
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Unit 3: Page 17 of 66
Mathematics III
Unit 3
2nd Edition
ii. In this problem, our exponent needs to represent the number of 5-hour
time periods that elapsed. If you represent 1 hour as 1/5 of a 5-hour time
period, how do you represent 2 hours? 3 hours? 10 hours? x hours?
Solution: 2 hours is 2/5 of a time period; 3 hours is 3/5; 10 hours is 2 5-hour time periods; x
hours is x/5 of a time period.
c. Using your last answer in part (b) as your exponent, write an exponential function
to model the amount of caffeine remaining in the blood stream x hours after the
peak level.
Solution: f(x) = 80(.5)x/5
d. How would use the function you wrote in (c) to answer the questions in (a)? Use
your function and check to see that you get the same answers as you did in part
(a). (Be careful with your fractional exponents when entering in the calculator.
Use parentheses.) If you need to, draw a line through your original answers in part
(a) and list your new answers.
Solution: f(0) = 80(.5)0/5 = 80; f(1) = 80(.5)1/5 = 69.644; f(2) = 80(.5)2/5 = 60.629; f(5) = 80(.5)1
= 40; f(10) = 80(.5)2 = 20
e. Determine the amount of caffeine remaining in the bloodstream 3 hours after the
peak level? What about 8 hours after peak level? 20 hours? (Think about how
many 5-hour intervals are in the number of hours you’re interested in.)
Solution: f(3) = 80(.5)3/5 = 53.78; f(8) = 80(.5)8/5 = 26.39; f(20) = 80(.5)4 = 5
f. Suppose the half-life of caffeine in the bloodstream was 3 hours instead of 5.
i. Write a function for this new half-life time.
Solution: f(x) = 80(.5)x/3
ii. Determine the amount of caffeine in the bloodstream after 1 hour, 2 hours,
5 hours, and 10 hours. (You need to consider how many 3-hour time
intervals are used in each time value.)
Solution: f(0) = 80(.5)0/3 = 80; f(1) = 80(.5)1/3 = 63.496; f(2) = 80(.5)2/3 = 50.397; f(5) = 80(.5)5/3
= 25.198; f(10) = 80(.5)10/3 = 7.937
iii. Which half-life time results in the caffeine being eliminated faster?
Explain why this makes sense.
Solution: Since the half-life is less in (f), caffeine is eliminated faster. It makes sense that the
answers in (f, ii) are smaller than those in (d). Because the caffeine is eliminated faster in (f),
there is less remaining in the bloodstream after the same number of hours than in (d).
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Mathematics III
Unit 3
2nd Edition
g. Graph both equations (from d and f) on graph paper. How are the graphs similar?
Different? What are the intercepts? When are they increasing/decreasing? Are
there asymptotes? If so, what are they? Do the graphs intersect? Where?
Solution: The graphs have
the same shape; the same
y-intercept (0, 80); they are
both always decreasing,
never increasing; they are
both asymptotic to the xaxis. They only intersect at
the y-intercept. After that,
because the second (halflife of 3 – red graph)
decreases faster than the
first, they will not intersect
again.
It makes sense, in thinking about time, that we
need all rational time values, e.g. 1/3 hour, 5/8
hour, etc. This raises the idea of rational
exponents, that is, computing values such as
33/4 or (1/2)7/3.
Dis cre te Data
100
Caffeine Concentration
Note that if we could only use integer
exponents; e.g. 1, 2, 3, etc; our graphs would
be discontinuous. We would have points (see
right), rather than the smooth, continuous curve
you graphed above.
80
60
40
20
0
0
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2
4
6
Hours
8
10
Mathematics III
Unit 3
2nd Edition
3. Rational Exponents. In previous courses, you learned about different types of numbers
and lots of rules of exponents.
a. What are integers? Rational numbers? Which set of numbers is a subset of the
other? Explain why this is true.
Solution: Integers are whole numbers and their opposites. Rational numbers are quotients of
integers, where the denominator is not equal to zero. Integers are a subset of rational
numbers. Every integer can be written as a rational number by writing it as a fraction of the
integer over 1.
b. Based on (a), what is the difference between integer exponents and rational
exponents?
Solution: An integer exponent means that the exponent is an integer value; a rational
exponent means that the exponent can be a rational number, including fractional or decimal
values.
c. Complete the following exponent rules. (If you don’t remember the rules from
your previous classes, try some examples to help you.)
For a > 0 and b > 0, and all values of m and n,
a0 = ___
a1 = ____
an = ________________________
(am)(an) = ________
(am)/(an) = ________
a-n = _______
(am)n = ________
(ab)m = ___________
(a/b)m = __________
If am = an, then m ___ n.
The same rules you use for integer exponents also apply to rational exponents.
Solution:
For a > 0 and b > 0, and all values of m and n,
an = a x a x … x a (a times itself n times)
a0 = 1
a1 = a
(am)(an) = am+n
(am)/(an) = am-n
a-n = 1/an
(am)n = amn
(ab)m = (am)(bm)
(a/b)m = (am)/(bm)
If am = an, then m = n.
d. You have previously learned that the nth root of a number x can be represented as
x1/n.
i. Using your rules of exponents, write another expression for (x1/n)m.
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Mathematics III
Unit 3
2nd Edition
Solution: xm/n
ii. Using your rules of exponents, write another expression for (xm)1/n.
Solution: xm/n
iii. What do you notice about the answers in (ii) and (iii)? What does this tell
you about rational exponents?
Solution: The answers are the same. So (x1/n)m = (xm)1/n = xm/n.
This leads us to the definition of rational exponents.
For a > 0, and integers m and n, with n > 0,
m
a n  am 
n

 
n
m
a ; am/n = (a1/n)m = (am)1/n .
e. Rewrite the following using simplified rational exponents.
7
i.
x3
1
ii.  
 x
5
iii.
 x
6
iv.
5
Solution:
i.
7
iv. 3
x3 = x3/7
1
x
5
1
ii.   = x5
 x
iii.
1
3
x5
 x = x
6
6/2
= x3
= x-5/3
f. Simplify each of the following. Show your steps. For example, 272/3 = (271/3)2 =
32 = 9.
i. 163/4
Solution:
ii. 363/2
i. 163/4 = (161/4)3 = 23 = 8
iii. 817/4 = (811/4)7 = 37 = 2187
iii. 817/4
ii. 363/2 = (361/2)3 = 63 = 216
Problems involving roots and rational exponents can sometimes be solved by rewriting
expressions or by using inverses.
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Unit 3: Page 21 of 66
Mathematics III
Unit 3
2nd Edition
g. Consider the caffeine situation from above. The half-life of caffeine in an adult’s
bloodstream is 5 hours. How much caffeine is remains in the bloodstream each
hour? Our original equation was f(x) = 80(.5)x/5. Use the rules of rational
exponents to rewrite the equation so that the answer to the above question is given
in the equation. What percent of the caffeine remains in the bloodstream each
hour?
Solution: f(x) = 80(.5)x/5 = 80(.51/5)x = 80(.8705505633)x; approximately 87% remains after an
hour.
To solve equations such as x3 = 27, we take the cube root of both sides. Alternately, we can raise
both sides of the equation to the 1/3 power. That is, we raise both sides of the equation to the
power that is the inverse (or reciprocal) of the power in the problem. To solve x3/2 = 27, we can
either square both sides and then take the cube root, we can take the cube root of both sides and
then square them, or we can raise both sides to the 2/3 power.
x3/2 = 27  (x3/2)2/3 = 272/3  x = (271/3)2  x = 32 = 9
h. Rewrite each of the following using rational exponents and use inverses to solve
for the variable. (You may need to use a calculator for some of them. Be careful!)
i.
Solution:
5
b 2
ii.
5
c3  4.2
iii.
4
1
1

d 5
i. b1/5 = 2  (b1/5)5 = 25  b = 32
ii. c3/5 = 4.2  (c3/5)5/3 = 4.25/3  c = 10.9323838
iii. d-1/4 = 1/5  (d-1/4)-4 = (1/5)-4  d = 54 = 625
Let’s look at some more problems that require the use of rational exponents.
4. Let’s use a calculator to model bacteria growth. Begin with 25 bacteria.
a. If the number of bacteria doubles each hour, how many bacteria are alive after 1
hour? 2 hours?
Solution: After 1 hour – 50; after 2 hours – 100
b. Complete the chart below.
Time (hours) 0
1
2 3 4 5 6
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Mathematics III
Unit 3
2nd Edition
25 50
Population
Solution:
Time (hours) 0
1
2
3
4
5
6
25 50 100 200 400 800 1600
Population
c. Write a function that represents the population of bacteria after x hours. (Check
that your function gives you the same answers you determined above. Think
about what if means if the base number is 1. What type of base number is needed
if the population is increasing?)
Solution: f(x) = 25(1 + 1)x = 25(2x). (100% of the population remains after an hour plus an
additional 100%.) The base number must be greater than 1 to indicate an increasing
population.
d. Use this expression to find the number of bacteria present after 7 1/2 and 15
hours.
Solution: f(7.5) = 25(27.5) = 4525.4834 bacteria; f(15) = 25(215) = 819200 bacteria
e. Suppose the initial population was 60 instead of 25. Write a function that
represents the population of bacteria after x hours. Find the population after 7 1/2
hours and 15 hours.
Solution: f(x) = 60(2x)
f(7.5) = 10861.16016
f(15) = 1966080 bacteria
f. Graph the functions in part (c) and (e). How are the graphs similar or different?
What are the intercepts? What do the intercepts indicate? When are they
increasing/decreasing? Are there asymptotes? If so, what are they? Do the graphs
intersect? Where?
Solution:
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Mathematics III
Unit 3
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Both graphs are increasing and are asymptotic to the x-axis. The two graphs do not intersect.
The intercept of the first is (0, 25) and the second is (0, 60). The intercepts indicate the
original population, the population at time 0 hours.
g. Revisit the graphs in problem (2). Compare with the graphs above. How are they
similar and different? How do the equations indicate if the graphs will be
increasing or decreasing?
Solution: The graphs in number 2 are always decreasing and the bases in the equations are
less than 1. The graphs above are always increasing and the bases in the equations are greater
than 1. All the graphs are asymptotic to the x-axis. All the graphs have y-intercepts that
correspond to the “original” number/concentration.
h. Consider the following: Begin with 25 bacteria. The number of bacteria doubles
every 4 hours. Write a function, using a rational exponent, for the number of
bacteria present after x hours.
Solution: f(x) = 25(2)x/4
i. Rewrite the function in (g), using the properties of exponents, so that the exponent
is an integer. What is the rate of growth of the bacteria each hour?
Solution: f(x) = 25(21/4)x = 25(1.189207115)x; 18.92% growth rate
j. If there are originally 25 bacteria, at what rate are they growing if the population
of the bacteria doubles in 5 hours? (Hint: Solve 50 = 25(1 + r)5.) What about if
the population triples in 5 hours? (Write equations and solve.)
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Mathematics III
Unit 3
2nd Edition
Solution: 50 = 25(1 + r)5  2 = (1 + r)5  21/5 = (1 + r)  21/5 – 1 = .148698355; 14.87%
If the population triples in 5 hours, we solve 75 = 25(1 + r)5. The rate is 24.57%.
k. If there are originally 25 bacteria and the population doubles each hour, how long
will it take the population to reach 100 bacteria? Solve the problem algebraically.
Solution: 100 = 25(2x)  4 = 2x  x = 2
l. If there are originally 60 bacteria and the population doubles each hour, how long
will it take the population to reach 100 bacteria? Explain how you solved the
problem. (Solving the problem algebraically will be addressed later in the unit.)
Solution: 100 = 60(2x)  5/3 = 2x  Using the intersecting graphs, we find that the
population reaches 100 approximately at .73696559 hours or 44 minutes, 13 seconds.
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Mathematics III
Unit 3
2nd Edition
Notes on The Population of Exponentia Learning Task
Although this task uses the same context as the launching task, it is not necessary to have
completed the launching task to complete this task.
Students are expected to recall knowledge of exponential functions from Math 2. The primary
standards addressed in this task are MM3A2c, e, f, and g. This task reviews MM2A2b, c, and
e: characteristics of exponential functions and modeling exponential functions; these ideas
are also stressed in MM3A2e, f, and g. It then establishes logarithmic functions as inverses of
exponential functions (element c) and engages students in investigating and determining the
characteristics of logarithmic functions (element e). Part 3 of the task familiarizes students
with using logarithms, particularly common logarithms, to solve exponential equations. All of
this is accomplished through the use of phenomena using exponential and logarithmic
functions (element g).
Please note that this unit does not address the properties of logarithms. This is intentional and
will be addressed in the next unit. The focus in this unit is on developing students’ conceptual
understanding of logarithms.
Supplies Needed:
 Patty paper or wax paper
 Graph paper
 Graphing calculators or computer graphing package
THE POPULATION OF EXPONENTIA
Upon landing on a nearly discovered planet, Exponentia, the astronauts discovered life on the
planet. Scientists named the creatures Viètians (vee-et-ee-ans), after the French mathematician
François Viète who led the way in developing our present system of notating exponents.
1. After observing the species for a number of years, NASA biologists determined that the
population was growing by 10% each year.
In Math 2, you studied exponential functions. Use that knowledge to answer the following about
the population of Exponentia.
a. If the estimated number of Viètians was 1 million in 1999 and their population is
increasing 10% a year, write an equation for the population of Exponentia, P, as a
function of the number of years, x, since 1999.
Solution: P(x) = 1(1 + 0.10)x = 1.1x, where population is measured in millions
b. What was the population in 2005? What would it be in 2015 if the population growth rate
remained the same?
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Mathematics III
Solution:
Unit 3
2nd Edition
P(2005 – 1999) = P(6) = 1.16 = 1.771561million = 1,771,561
P(2015 – 1999) = P(16) = 1.11.6 = 4.594972986 million = 4,594,972.986
c. Graph the function in part (a) on the interval -15 ≤ t ≤ 15.
Solution:
i. What are the domain and range of the entire function?
Domain: ______________________
Solution: Domain: All real numbers
Range: _____________________
Range: y > 0
ii. Identify any asymptotes and intercepts. What do these characteristics tell
us about the population on Exponentia from 1999 to 2009?
Asymptotes: _____________________
Intercepts: _____________________
Meanings: _________________________________________________________
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Unit 3: Page 27 of 66
Mathematics III
Unit 3
2nd Edition
Solution:
Asymptotes: y = 0
Intercepts: (0, 1)
Meanings: In this context, the asymptote tells us that the population was never 0. The
intercept indicates that at time 0 (1999), the population was 1 million.
iii. When is the function increasing / decreasing? What does this tell us about
the population on Exponentia from 1999 to 2009?
Solution: The function is always increasing. The population was increasing during the time
period.
d. Why would it be useful to have an equation that expresses the number of years since
1999 as a function of the population of Exponentia? That is, why might you want to be
able to determine the year if you knew the population of Exponentia?
Solution: (Answers will vary) Biologists might want to know in what year the population was 1
or in what year the population would equal the population of Georgia, etc.
e. How can you use the graph in part (c) to determine the year in which the population
reaches 2 million Viètians?
Solution: We could see where the line y = 2 intersects the graph in part (c).
f. Write an equation, based on your function in part (a), for how you would find the year the
population reaches 2 million. Use your graphing calculator to determine the solution.
Solution: 2 = 1.1x; x = 7.27 years
We want to be able to solve the equation in part (f) algebraically. To solve x3= 27 for x, we apply
the inverse of cubing, taking the cube root, to both sides of the equation. To solve 2x = 10 for x,
we apply the inverse of multiplying by x, dividing by 2, to both sides of the equation. Similarly,
to solve the equation in (f), we need to apply the inverse of raising a value to the x power; we
need to find the inverse of exponential functions.
After the following exploration, we will return to this problem and solve it algebraically.
2. Graphs of exponential functions and their inverses.
a. Graph the function f(x) = ax, where a = ½, ¾, 2, 5, and 3/2, on your graph paper. (You
will have a total of 5 graphs.)
Solution: All five graphs are included in the following picture. It might be useful for students
to graph them separately but using the same increments on their graphs.
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Mathematics III
Unit 3
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a = ½ (baby blue graph); a = ¾ (blue graph); a = 3/2 (purple graph); a = 2 (red graph); a = 5
(green graph)
b. Using the above graphs, answer the following.
For a > 0, a  1,
Domain: _________________
Range: ___________________
Asymptote: _______________
Intercept: _________________
Increases when _________________; Decreases when _______________
Solution:
For a > 0, a  1,
Domain: All real numbers
Range: y > 0
Asymptote: y = 0
Intercept: (0, 1)
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Mathematics III
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Increases when a > 1; Decreases when 0 < a < 1
c. Graph f(x) = 2x. List at least two points that are on the graph of f(x).
Solution: Two points will vary. Ideally, one would be the intercept, (0, 1). Others could include
(1, 2), (2, 4), (3, 8), etc.
How is each a transformation of f(x)? For example, if g(x) = 2x+4, it is a transformation of f(x) 4
units to the left; g(x) = f(x+4).
Explain which, if any, of the characteristics listed in part (b) will change.
Also, for each of the points you listed above, determine the location of those points after the
transformations in the new equations.
(Think about how changes in the equations transform the graphs of the equations. If necessary,
graph each function, along with f(x), to answer this question.)
i. g(x) = 2x – 3
ii. h(x) = 22x
iii. j(x) = 3(2x)
iv. k(x) = 2-x
v. l(x) = -2x
vi. m(x) = 2x+3
vii. n(x) = 2x- 1
Solution:
i. g(x) = f(x – 3). The graph will be shifted 3 units to the right. The domain, range,
and asymptote will remain the same; the intercept will be different. Instead of
the points (0, 1) and (1, 2), two of the points on the graph will be (3, 1) and (4,
2).
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Mathematics III
Unit 3
2nd Edition
ii. h(x) = f(2x). This is a horizontal stretch by ½ (or a horizontal shrink by a factor
of 2). All of the characteristics from (b) would remain the same. (0, 1)  (0, 1);
(1, 2)  (1/2, 2) or (1, 4), depending on how it is viewed.
iii. j(x) = 3f(x). j(x) is a vertical stretch of f(x) by a scale factor of 3. The domain,
range, and asymptote remain the same. The intercept, however, changes. (0, 1)
 (0, 3); (1, 2)  (1, 6).
iv. k(x) = f(-x). A negative in front of the independent variable changes increasing
/ decreasing nature of the exponential function. In this case, because the
function was increasing, the function becomes decreasing. This can be seen in
the following: g(x) = 2-x = (1/2)x. As stated in (b), whenever the base is between
0 and 1, the function is decreasing. However, the domain, range, intercept, and
asymptote do not change. (0, 1)  (0, 1); (1, 2)  (1, ½) or (-1, 2), depending
on how it is viewed.
v. l(x) = - f(x), so l(x) is a reflection of f(x) over the x-axis. The domain is still all
real numbers and the asymptote remains the same. However, the function was
increasing, but it is now decreasing; the intercept has changed. (0, 1)  (0, -1);
(1, 2)  (1, -2).
vi. m(x) = f(x) + 3. This is a shift of the graph up 3 units. The domain remains the
same, but all other characteristics change. The range is y > 3; the asymptote is y
= 3; the intercept is (0, 4). (0, 1)  (0, 4); (1, 2)  (1, 6)
vii. n(x) = f(x) – 1. This is a shift of the graph of f(x) down 1 unit. The domain
remains the same, but all other characteristics change. The range is y > -1; the
asymptote is y = -1; and (0, 0) is both the x- and y-intercept. (0, 1)  (0, 0); (1,
2)  (1, 1).
d. We need to determine the inverse of an exponential function to algebraically complete
problem 1 and to determine x-intercepts of more interesting exponential functions. List
everything you know about graphs of inverse functions.
Solution: Answers will vary but should include that the domain and range of inverse functions
are interchanged; their graphs are reflections over the line y = x; they are found algebraically
(in functions written in terms of x and y) by interchanging the variables and solve for the new
dependent variable; horizontal and vertical line tests (and how they determine if the inverses
are functions – one-to-one functions have functional inverses); and that the composition of a
function and its inverse yields f(g(x)) = x.
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Mathematics III
Unit 3
2nd Edition
e. Paper folding: Using a piece of patty paper or wax paper, trace an accurate set of
coordinate axes on the patty paper. Include all four quadrants.
i. Draw an accurate graph of f(x) = 2x on the patty paper. Also draw the line y = x.
ii. Fold the paper along the line y = x.
iii. Trace the curve, as you can see it, on the outside of the paper.
iv. Open the paper and trace the outside trace on the inside so that you have both the
graph of the exponential function and the graph of its inverse on the front of the
patty paper.
v. Using the graph on your patty paper, make conjectures about the following
characteristics of the inverse of the exponential function.
Domain: _________________
Range: ____________________
Asymptote: _______________
Intercept: __________________
Solution: (Answers will vary based on how well the students did their paper folding.)
f. Using your lists from parts (b) and (e) and your conjectures in part (f), list the
characteristics of the inverse of the exponential function. Draw the graph of this new
function.
Domain: _________________
Range: ____________________
Asymptote: _______________
Intercept: __________________
Graph:
Solution: (Here, the idea is that students are using their knowledge of inverse functions,
confirmed by their paper folding exercise, to visualize and understand the graph of the inverse
of the exponential.)
Domain: x > 0
Range: All real numbers
Asymptote: x = 0
Intercept: (1, 0)
Graph:
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Mathematics III
Unit 3
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3. A logarithmic function is the inverse function of an exponential function.
a. In an exponential function, the input or independent variable is the exponent and
the output or dependent variable is the value when the base is raised to an
exponent. For example, in f(x) = 3x, if x = 4, then f(x) = 34 = 81. What are the
independent and dependent variables in a logarithmic function? (Explain in
words, similar to what is done for exponents.)
Independent: _____________________________________
Dependent: ______________________________________
Solution: Because exponentials and logarithms are inverses, the independent and dependent
variables are interchanged. So the independent is the value when a base is raised to a power
and the dependent in the exponent.
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Unit 3: Page 33 of 66
Mathematics III
Unit 3
2nd Edition
b. Logarithmic notation: To find the equation of inverses, we interchange x and y
and solve for y.
y = bx  x = by
However, we need a method for solving for y. Just as when we need to take the inverse of
a power function we use nth roots (square roots, cube roots, etc.), we need to introduce a
new notation for the inverse of an exponential function.
For y and b both positive and b  1,
y = bx if and only if x = log b y
This is read as “x equals the log base b of y” or “x equals the base-b logarithm of y.”
Note that the “base” is the “base” in both expressions. A logarithm of a positive number y
is the exponent x you write y as a power of a base b.
Let’s look at a few examples:
102 = 100 can be written as log10100 = 2. Notice that the answer to the logarithm is the
exponent from the first equation.
Evaluate log464. This question asks for the value that when 4 is raised to that power is 64.
In other words, 4 to what power equals 64? ________
Solution: 3
Similarly, consider the following problem: logx144 = 2. Here, we consider the base that
when raised to the second power, squared, yields 144. So x2 = 144. Given that the base
must be positive, what is the value of x? ______
Solution: 12
c. Practice with logarithms:
i. Write each exponential as a logarithm or logarithm as an exponential.
1. log10(1/100) = -2
2. 53 = 125
3. log232 = 5
Solution:
1. 10-2 = 1/100
3. 25 = 32
2. log5125 = 3
ii. Evaluate each logarithm.
1. log10(0.1)
2. log381
3. log2(1/16)
4. log981
Solution:
1. -1
2. 4
3. -4
4. 2
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iii. Solve each logarithmic equation for x.
1. logx36 = 2
2. log44 = x
3. log71 = x
4. log8x = 3
5. log5(3x + 1) = 2
6. log6(4x – 7) = 0
Solution:
1. x = 6
4. x = 512
2. x = 1
5. x = 8
3. x = 0
6. x = 2
d. Understanding the logarithm definition:
For y and b both positive and b  1,
y = bx if and only if x = log b y
i. The statement of the relationship between exponents and logarithms states
that b  1. Explain why b  1 if y = bx is an exponential function.
Solution: If b = 1, then y = 1x. 1 raised to any power is 1. This would yield a horizontal line of
y = 1, not an exponential function.
ii. The statement of the relationship between exponents and logarithms also
states that y and b are both positive. Explain why the base, b, must be
positive for y = bx to be an exponential function. Then explain why y must
be positive if b is positive. Be sure to consider both positive and negative
values of x.
Solution: Consider y = bx. If b < 0, the values of y oscillate between positive and negative
values. This is not an exponential function. So b > 0. Because b > 0, regardless of the power to
which b is raised, positive or negative, the output will always be positive.
e. Common logarithms: In logarithms, just as with exponential expressions, any
positive number can be a base. However, the base of our decimal counting
system, 10, is called the common base. Logarithms which use 10 for the base are
called common logarithms and are notated simply as log x. It is not necessary to
write the base. The log button on your calculator computes common logarithms.
Solve the following: log 1000 = x.
Solution: x = 3
Understanding common logarithms can help solve more complex exponential
equations.
Consider the following: Solve for x. 10x = 250
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We know that 102 = 100 and 103 = 1000, so x should be between 2 and 3. Rewriting
as a common logarithm can help solve the problem.
log 250 = x
Using your calculator, find x. x = _______ Solution: x = 2.3979
f. Solve each of the following for x using common logarithms.
i. 10x = 15
ii. 10x = 0.3458
iii. 3(10x) = 2345
iv. -2(10x) = -6538
Solution:
1. x  1.1761
2. x  -0.4612
3. x  2.8930
4. x  3.5144
4. Solving exponential equations: At numerous times in this unit, we have only been able
to solve exponential equations using tables and graphs (e.g., How long does it take, #1f-g;
How long does it take, #4l; Population of Exponentia, #1f). We can now use our
understandings of common logarithms to solve these kinds of problems.1
a. Consider the following problem: A bacteria culture in a Petri dish was begun with
a single bacterium. Once the population reaches 100, the bacteria begin dying.
How long does it take the population of the bacteria in the Petri dish to reach 100?
Here’s the equation we need to solve: 2x = 100
We know that 26 = 64 and 27 = 128, so the answer must be between 6 and 7.
We can rewrite each side of the equation as a power with a base of 10. That is, 10a
= 2 and 10b = 100. So a  0.3010 and b = 2.
Explain how we know the values of a and b.
Solution: Similar to above, 10a = 2 can be rewritten as a = log 2. This value can be found on
the calculator. The value for b is clear due to the numbers involved.
Now, replacing 2 with 10.301 and 100 with 102, we have the following equation:
(10.301)x = 102
10.301x = 102
Power-to-power property
0.301x = 2
Common base property
x  6.6439
1
Adapted from Discovering Advanced Algebra: An Investigative Approach from Key Curriculum Press, 2004.
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Explain why you might be confident that this is the correct answer.
Solution: We established that the answer would be between 6 and 7. This answer is, indeed,
between those numbers.
b. Solve each of the following equations, taken from earlier explorations, using the
above method.
i. 2 = 1.1x
1. 10a = 2 and 10b = 1.1. Find a and b. a = _____ b = _____
2. Use substitution to write an equation in terms of powers of 10.
________________
3. Use the power-to-power and common base properties.
___________________
4. Solve for x. x = ______________
ii. 300(.8)x = 10
iii. 100 = 60(2x)
Solution:
i.
1. a = 0.301; b = 0.0414
2. (10.301)x = 10.0414
3. .301x = .0414
4. x  .1375
ii. .8x = 1/30  10-.0969x = 10-1.4771  -.0969x = -1.4771  x  15.2436
iii. 2x =5/3  10.301x = 10.2218  .301x =.2218  x  0.7369
c. This method can also be used to determine x-intercepts of exponential functions.
Consider the equations in 2c of this task.
i. Which functions would have x-intercepts if you were to graph them? How
do you know without graphing them?
Solution: Only the graph of the last function, n(x), would have an x-intercept. The standard
exponential does not have an intercept, so shifts left or right, stretches, and reflections (in
isolation) would not create an x-intercept. A vertical shift down could, however, result in an xintercept.
ii. Select one of functions that would have an x-intercept. Determine, using
the method above, the x-intercept.
Solution: The only option is n(x) = 2x – 1. To find the x-intercept, set the function equal to
zero and solve for x. 0 = 2x – 1  2x = 1  10.301x = 100  .301x = 0  x = 0. (Discuss how to
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solve this problem using mental math—because 2x = 1, x must equal 0—as well as the method
above.)
iii. Select one of functions that would have an x-intercept. Start solving for the
x-intercept. How does your answer confirm that the graph of the function
has no x-intercept?
Solution: (Answers will vary. Only the first one is worked out here.)
g(x) = 2x-3  0 = 2x -3  There is no power of 10 that equals 0, so there is no solution for x.
Therefore, the graph of this function does not have an x-intercept.
5. Graphs of logarithmic functions: If f(x) =bx, and the inverse of an exponential function
with respect to x is a logarithmic function with respect to x. So f -1(x) = logbx.
In the paper folding activity above, you first graphed f(x) = 2x and then drew in the graph of the
inverse function. What is the equation of f -1(x)? _________________
Solution: f -1(x) = log2x
We want to make explicit the characteristics that are common to ALL general logarithmic
functions, regardless of the base. To do this, we need to look at graphs of different logarithmic
functions.
a. Let’s graph y = log5x by hand.
i. How can you use the fact that y = log5x is the inverse of y = 5x to graph y =
log5x?
Solution: Answers may vary. One possibility is to make a table of values for y = 5x and
interchange the domain and range to graph y = log5x. Another idea is to graph the exponential
function, reflect over the line y = x, and trace the reflection.
ii. Make a table of values to help you graph y = log5x. (Don’t forget to use xvalues between 0 and 1. Consider y = 5x in making your table.)
Solution: (Possible table given.)
x
1
5^(1/4)
5
5
25
125
1/5
y = log5x
0
¼
½
1
2
3
-1
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-2
iii. Referring to your graph in 2f and your table above, draw the graph of y =
log5x.
Solution:
iv. Use the same procedure to draw the graph of y =log3x.
Solution: (Possible table given.)
X
y = log3x
1
0
3^(1/4)
¼
½
3
3
1
9
2
27
3
1/3
-1
1/9
-2
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v. Compare the graphs of y = log5x and y = log3x. Which one increases faster?
How can you tell?
Solution: The graph of y = log3x increases faster than the graph of y = log5x. One way to
see this is that the point (1, 0) is on both graphs. And each graph has a point with a ycoordinate of 1. It only takes a horizontal distance of 2 (from the point (1, 0)) for y = log3x
to reach a y-value of 1, but it takes a horizontal distance of 4 for y = log5x to reach a yvalue of 1.
b. To better understand the characteristics of all logarithmic functions, use your the
list feature on your graphing calculator or computer graphics tool to complete the
following table.
y = log5x
y = log3x
y = log(1/3)x
y = log (1/5)x
y = log5x
y = log3x
y = log(1/3)x
y = log(1/5)x
Domain
(0, ) or x > 0
(0, )
(0, )
(0, )
Range
All real numbers
All real numbers
All real numbers
All real numbers
Intercept
(1, 0)
(1, 0)
(1, 0)
(1, 0)
Asymptote
x=0
x=0
x=0
x=0
Zeros
(1, 0)
(1, 0)
(1, 0)
(1, 0)
Increasing /
increasing
Increasing
Decreasing
Decreasing
Domain
Range
Intercept
Asymptote
Zeros
Increasing /
Decreasing
Solution:
Decreasing
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c. Using the information above, when do logarithmic functions increase? When do
they decrease?
Solution: Consider y = logbx. Logarithmic functions decrease when 0 < b < 1 and increase
when b > 1.
Give an example of a function that increases faster than the increasing functions
in part (b). Explain how you determined your function.
Solution: One possible answer: y = log2 x. The smaller the value of the base, b > 1, the faster
the graph increases.
Which of the above graphs decreased fastest?
Solution: The function with a base of 1/3 decreased faster than that with a base of 1/5.
Give an example of a function that would decrease faster than any of the ones
above. Explain how you determined your function.
Solution: One possible answer: y = log(1/2)x. The larger the value of the base, 0 < b < 1, the
faster the graph decreases.
d. The definition of a logarithm states that x must be positive in y = logbx.
Accordingly, our domain is that x is positive. What happens when we transform
the equation?
1. Consider y = log (x – 3). This is a transformation of the logarithmic graph to
the right 3 units. Sketch the graph of this function.
Solution:
What is the new domain?
What is the new asymptote?
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What is the new intercept?
How does the transformation relate to the changes above?
Solution: The domain is x > 3. The asymptote is x = 3. The new intercept is (4, 1). The
transformation was a horizontal shift to the right 3 units. Each of these attributes shifted to
the right 3 units.
2. For more interesting functions like y = log4(3x – 5), we can use the same idea
to determine graph attributes.
The domain will be determined by solving the inequality 3x – 5 > 0. The
asymptote will be determined by solving the equation 3x – 5 = 0.
Find the domain. __________
Solution: Domain x > 5/3
Find the asymptote. ______________
Asymptote x = 5/3
Generally, how do you find x-intercepts? Find the x-intercept for this function by
solving the associated logarithmic equation..
Solution: An x-intercept is where the graph intersects the x-axis, or where the y-coordinate is
equal to 0. Setting y = 0, we get the following:
0 = log4(3x – 5)

40 = 3x – 5

1 = 3x – 5

x=2
So the x-intercept is (2, 0).
3. Find the domain, asymptote, x-intercept, and y-intercept (if applicable) of
y = log5(-3x + 8).
Domain: ____________
Asymptote: ______________
x-intercept: __________
y-intercept: ______________
Solution:
Domain: x < 8/3
x-intercept: (3, 0)
Asymptote: x = 8/3
y-intercept: (0, 1.292)
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Notes on Modeling Natural Phenomena on Earth Learning Task
Although this task uses the same context as the launching task, it is not necessary to have
completed the launching task to complete this task.
The primary standards addressed in this task are MM3A2f and g, transformations of graphs
and exploring real phenomena related to exponential and logarithmic functions. This task
addresses the definition of the natural logarithm through problems concerning the natural
exponential function. Additionally, in solving problems related to real life phenomenon, f(x) =
ex and f(x) = ln x will be graphed. Some basic ideas of e will be reviewed in the course of the
task. The task is primarily focused on graphing and applying logarithms to situations.
Supplies Needed:
 Graph paper
 Calculators
MODELING NATURAL PHENOMENA ON EARTH
NASA has decided to send astronauts to explore Exponentia, a new planet recently discovered
by NASA astronomers. Both the astronauts and the scientists realized that many preparations
must be made before the trip. For one thing, the astronauts need to get their lives, including their
financial situations, in order. And, in preparation for comparing the characteristics of Exponentia
with natural phenomena on Earth, the scientists need to revisit, as two examples, how sound
travels and how earthquakes are measured.
1. NASA must ensure that the astronauts have sufficient drinking water while they are
traveling to Exponentia. Part of the scientists’ jobs is to test the pH of the water that is to
be stored on the space shuttle.
pH is a measure of number of hydrogen ions, [H+], in a substance and measures how
acidic or basic (alkaline) a solution is. When measuring pH, we often talk of measuring
the acidity of a solution, assigning pH numbers between 0 and 14 that show relative
acidity. A solution that has a pH less than 7 is acidic and a solution that has a pH greater
than 7 is basic. A solution with a pH of 7 is neutral.
[H+] is usually a very large or very small number. Because logarithms can be used to
represent very large or very small numbers, we can use logarithms for this conversion.
The pH is the negative logarithm of the concentration of free hydrogen ions, measured in
moles per liter (moles/L). The conversion formula is pH = -log [H+].2
2
Information about acidity was gathered from a variety of web resources, including Wilkes University (www.waterresearch.net/ph.htm) and the Georgia Cooperative Extension Service
(http://pubs.caes.uga.edu/caespubs/pubs/PDF/B1242-3.pdf).
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a. Consider the general common logarithmic function, f(x) = log x. How will the
graph of the pH conversion function, g(x) = - log x, differ from the graph of f(x)?
Specifically, how, if at all, are the domain, range, intercepts, asymptotes,
increasing/decreasing changed? What kind of graphical transformation is this?
Solution: The conversion function is a vertical reflection of f(x) over the x-axis. The domain,
range, intercept, and asymptote will remain the same. However, the graph of g(x) is
decreasing.
b. If water has a pH of 5, what is the concentration of hydrogen ions?
pH = - log [H+] so we have 5 = - log [H+]. Rewrite this equation as an exponential
equation. Then solve the equation.
Solution: -5 = log [H+]  10-5 = [H+]. So the concentration is 10-5moles/L.
What if the water has a pH of 7? How does the concentration of hydrogen ions
compare? Explain why your answers make sense.
Solution: If the pH is 7, the concentration is 10-7 moles/L. Water with a pH of 5 has a
hydrogen ion concentration that is 100 times that of water with a pH of 7. This makes sense
because the base is 10 and it is a negative relationship. Each decrease in 1 in the pH
corresponds to a 10-fold increase in the hydrogen concentration.
c. The normal [H+] range for drinking water is between approximately 10-6 and 3.16
x 10-9. What is the approximate pH range?
Solution: The approximate pH range is between 6.0 and 8.5.
d. The rover that was sent to test out the surface of Exponentia brought back a
sample of a substance whose concentration of hydrogen ions measured 10-4. Is
this more or less acidic than our water? Explain. (Find the pH level of the
substance.)
Solution: The pH of this substance is 4. The lower the pH, the more acidic it is. Therefore, this
substance is more acidic than our drinking water.
2. The explorer rover sent to Exponentia experienced planetary tremors similar to what we
call earthquakes. The NASA scientists decided to train the astronauts to understand
earthquake measurements in case they experienced these tremors while on Exponentia.
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a. In discussing the relative size of earthquakes, for each increase in one point on the
Richter scale, the relative size increases ten-fold.3
Complete the following table that shows the correspondence of relative size and
Richter scale number.
Richter Scale Number
Relative Size
1
10
2
3
…
…
6
Solution:
Richter Scale Number
Relative Size
1
10
2
100
3
1000
…
…
6
1,000,000
b. Write a logarithmic equation for the relationship between Richter number and
relative size of the earthquake. Let relative size, s, be your independent variable
and Richter scale number, r, be the dependent variable. What is the name for this
special type of logarithm?
Solution: r = log s. This is the common logarithm.
3
Some of the content for this question derives from Integrated Mathematics 3 by McDougal-Littell (2002) and from
the U.S. Geological Survey website (http://earthquake.usgs.gov).
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c. The planetary rover experienced earthquakes of magnitudes 2 and 8. We want to
know the relationship between the relative size of an earthquake that measures 8
on the Richter scale and one that measures 2.
What is the relative size of an earthquake of magnitude 2? Write this as a power
of 10. ________
What is the relative size of an earthquake of magnitude 8? Write this as a power
of 10. ________
What is the relationship between the relative sizes? (Hint: Subtract the
exponents.)
Solution: 102; 108; The second is 106 times greater than the first.
d. In 2002, an earthquake of magnitude 7.9, one of the largest on U.S. land, occurred
in the Denali National Park in Alaska. What was the relative size of this
earthquake? (You will need to rewrite the logarithmic equation as an exponential
equation.)
Solution: 7.9 = log s  107.9 = s  s = 79,432,823.47
e. On April 29, 2003, an earthquake in Fort Payne, Alabama was felt by many
residents of northern Georgia. The magnitude was 4.6. How does the relative size
of the Alabama earthquake compare with the relative size of the Denali
earthquake?
Solution: 4.6 = log s  104.6 = s  s = 39,810.71706. The Denali earthquake was 103.3 times
the size of the Alabama earthquake.
3. Rather than discuss relative size of an earthquake, we often prefer to discuss the amount
of energy released by an earthquake. A formula that relates the number on a Richter scale
to the energy of an earthquake is r = 0.67 log E – 7.6, where r is the number on the
Richter scale and E is the energy in ergs.
a. Consider the general form of a logarithmic function, f(x) = log x.
How will the scale factor of 0.67, in the equation above, alter the graph of f(x)?
What is the name of this transformation?
How will the –7.6 alter the graph of f(x)?
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What is the name of this transformation?
Solution: The scale factor of 0.67 will make the graph increase slower. It is a vertical
compression. The – 7.6 will shift the graph down 7.6 units. This is a downward vertical shift.
b. Determine the domain, range, intercepts, and asymptotes for the graph of
r = 0.67 log E – 7.6. Sketch the graph.
Recall that for the domain and asymptote, we need to consider the value for which
the logarithm is being taken. In this case, we are taking the logarithm of E.
Domain: ________________
Range: _____________
Asymptote: _______________
x-intercept: _____________
No option is provided above for a y-intercept. Why not? What type of
transformations might result in the graphs having y-intercepts? (Investigate on
your calculator if you would like.)
Solution:
Domain: E > 0
Range: r is all real numbers
(however, we only use r > 0)
Asymptote: E = 0
x-intercept: (2.2 x 1011, 0)
No option is given for a y-intercept because these transformations would not
create a y-intercept. A horizontal reflection and horizontal shift to the left are
two transformations that might result in y-intercepts.
c. What is the Richter number of an earthquake that releases 3.9 x 1015 ergs of
energy? (Be careful when inputting this into the calculator.)
r = 0.67 log E – 7.6
r = 0.67 log (3.9 x 1015) – 7.6
r = ________
Solution: r = 2.846
d. How much energy was released by the 2002 Denali earthquake? By the 2003
Alabama earthquake?
Denali earthquake: r = 0.67 log E – 7.6
7.9 = 0.67 log E – 7.6
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Solve this equation for E by first solving for log E and then rewriting the equation
in exponential form. E = ____________
Solution: log E = 15.5/.67  1023.12 = E  1.36 x 1023 ergs of energy
Alabama earthquake: r = 0.67 log E – 7.6
4.6 = 0.67 log E – 7.6
E = ____________
Solution: log E = 12.2/.67  1018.21 = E  1.62 x 1018 ergs of energy
4. Scientists use carbon-dating to determine the ages of carbon-based substances. The
isotope Carbon-14 (C14) is widely used in radiocarbon dating. This form of carbon is
formed when plants absorb atmospheric carbon dioxide into their organic material during
photosynthesis. After plants die, no more C14 is formed and the C14 in the material
declines exponentially.4
a. Willard Libby was recognized with the Nobel Prize in Chemistry in 1960 for
leading a team to develop a method of using carbon-14 dating to determine the
age of carbon-based matter. Initially, circa 1949, Libby and associates believed
the half-life of the C14 isotope, the amount of time it takes for half of the C14 to
decay, to be approximately 556830 years. This is known as the Libby half-life.
Using 5568 as the value of the half-life, find an equation that models the
percentage of C14 in carbon-based material, such as a plant.
Let’s assume that we start with a plant that contains 100% of its C14, at the
moment it is cut.
If the initial amount is 100%, how much will remain at the end of one half-life?
______ In the equation y = abx, a = the initial amount. Will the b be greater than
1 or less than 1? ____ How do you know?
Solution: 50%; b < 1; The amount of C14 is decreasing so the base should be less than one to
indicate this decrease.
Using y = abx, substitute in the values you know, using the amount of time for the
exponent. (The only unknown is the base, b.)
Solution: 50 = 100(b)5568
4
Credits for information on carbon-14 dating: www.c14dating.com/int.html
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Now use a graphing calculator to find the value of b by allowing Y1 to equal the
left-hand side of your equation and Y2 to equal the right-hand side of your
equation. b = _________
Rewrite the equation y = abx with the values for a and b. This is the equation for
determining how long a plant has been decaying based on the amount of C14
remaining in the plant.
Solution: b = .99987552; y = 100(.99987552)x
b. Later measurements led to the realization that the Libby half-life was too low. In
actuality, the half-life of C14 is approximately 1.03 times the Libby half-life. This
is called the Cambridge half-life and is the accepted value for the half-life of the
C14 isotope.
Find the value of the Cambridge half-life. ___________
The actual value of the Cambridge half-life is 573040 years.
Will the base in the exponential equation that uses the Cambridge half-life be
smaller or larger than the base in the equation above? Explain.
Using 5730 as the value of the half-life, find an equation that models the
percentage of C14 in a plant. (Use the same procedure as above.)
Solution: 5568(1.03) = 5735.04 years; The base will be bigger because the C14 decays faster
than originally thought; b = .999879; y = 100(.999879)x
c. How are the graphs of the functions in (a) and (b) similar? How are they
different?
Solution: The graphs will be nearly identical with the same domain, range, intercept, and
asymptote. The rate of increase is slightly different. The function in (b) increases slightly
faster than the function in (a).
d. A plant contains 64.47% of its original carbon-14. According to this information,
approximately how long ago did it die? (3627.6 years)
Using the equation from (b), substitute 64.47 for the ending amount, y.
We need to solve for x. Divide both sides by 100 and then use common
logarithms to rewrite each side as 10 to a power and solve.
(Note: This method is explained in The Planet of Exponentia #3g.)
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Mathematics III
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Solution: 64.47 = 100(.999879)x; .6447 = .999879x; 10-.1906= 10(-.00005)x; 3812 years ago. (Less
rounding would result in an answer closer to 3627.6 years old.)
e. The planetary rover sent to Exponentia brought back a number of carbon-based
plants. Assuming that the plants died the moment they were gathered by the rover
in 2002, how much of the C14 would remain in 2222?
Use your equation in (b) to answer this question.
Solution: y = 100(.999879)20 = 99.76, so 99.76% remains after 20 years.
5. Before leaving for the exploration trip, one of the astronauts, Natalie Napier, met with her
accountant to make sure that her money was earning as much as possible while she was
gone.
a. In Math 2, you learned how to calculate investments that are compounded
different numbers of time, as well as compounded continuously. State those
formulas. Explicitly define the parameters (variables) in the formulas.
Solution: A = P0(1 + r/n)nt, where A is the ending amount, Po is the initial amount, r is the
annual interest rate, n is the number of times compounded per year, and t is the number of
years. The formula for compounded continuously is A = P0ert, with the same parameter
definitions.
b. The compounded continuously formula employs the transcendental number e.
Recall that a transcendental number is a number that is not the solution of a nonzero polynomial with rational coefficients. That means that e cannot be the
solution of an equation such as 2x2 + 3x – 4 = 0. Can you think of another
transcendental number? ______
Solution: 
Let’s review how we can estimate the value of e. Using the table feature on your
calculator (in the “ask” mode) and the compounded interest formula (not the
compounded continuously formula), complete the following chart. Assume that
you invest $1 at a 100% annual interest rate.
What is the equation, leaving only n and the ending amount as the unknowns?
_______________
Solution: A = 1(1 + 1/n)n
Input this at Y1. Change the “Tblset” to start at 1 and set the independent variable
option as “ask.” Now go to the table and type the n values for which you want the
dollar value.
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Mathematics III
Unit 3
Frequency of
Compounding
Number of times
compounded in a
year
2nd Edition
Formula with values
inputted
Dollar value at the
end of 1 year
Annual
Semiannual
Quarterly
Monthly
Weekly
Daily
Hourly
Every Minute
Every Second
Solution:
Frequency of
Compounding
Annual
Semiannual
Quarterly
Monthly
Weekly
Daily
Hourly
Every Minute
Every Second
Number of times
compounded in a
year
n=1
n=2
n=4
n = 12
n = 52
n = 365
n = 8760
n = 525600
n = 31536000
Formula with values
inputted
A = 1(1 + 1/1)
A = 1(1 + 1/2)2
A = 1(1 + 1/4)4
A = 1(1 + 1/12)12
A = 1(1 + 1/52)52
A = 1(1 + 1/365)365
A = 1(1 + 1/8760)8760
A = 1(1 + 1/525600)525600
A = 1(1 + 1/3153600)3153600
Dollar value at
the end of 1
year
2
2.25
2.44140625
2.61303529022
2.69255969544
2.71456748202
2.71812669063
2.7182792154
2.71828247254
Now, use your calculator to find the value of e. ______________________
What do you notice when looking at the table and looking at the value for e? That
is, how could we estimate the value of e?
Solution: e = 2.718281828. As the value of n increases, the dollar amount gets closer to the
value of e. If we used a really large n, we could get an even better estimate for e.
c. Assume that Natalie has $10,000 to invest. Complete the following chart to show
how much she would earn if her money was invested in each of the specified
accounts for 10 years.
Frequency of
Compounding
1. Quarterly
Annual Interest
Rate
3.65%
Formula with
Values Inputted
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Amount After
10 Years
Mathematics III
Unit 3
2. Monthly
3. Daily
4. Continuously
3.65%
3.6%
3.6%
Frequency of
Compounding
1. Quarterly
Annual Interest
Rate
3.65%
2nd Edition
Solution:
2. Monthly
3.65%
3. Daily
3.6%
4. Continuously
3.6%
Formula with
Values Inputted
A = 10,000(1 +
.0365/4)^40
A = 10,000(1 +
.0365/12)^(120)
A = 10,000(1 +
.036/365)^(3650)
A=
10,000e^(.036*10)
Amount After
10 Years
$14,381.32
$14,397.16
$14,333.04
$14,333.29
Which account would you suggest for Natalie? ________________
Solution: Natalie would make the most money with the account in which the interest is
compounded monthly at 3.65% per year.
d. Natalie is particularly interested in how long it will take her money to double.
Consider the first option. If her money doubles, she will have $20,000. So the
equation we want to solve is
20,000 = 10,000(1 + .0365/4)(4t)  2 = 1(1 + .0365/4)(4t)  2 = (1.009125)4t.
Remember that to solve some exponential equations, we must employ the inverse
operation, logarithms. Recall that we can write each base as 10 to a power because a
logarithm is the exponent when a base is raised to a power.
2 = 10a  log 2 = a  a = ___________
1.009125 = 10b  log 1.009125 = b  b = _____________
2 = (1.009125)4t  10.____ = 10____*4t  ______ =_____t  t = _______ years
Solution: a = .301; b = .0039449655; 10.301 = 10.0039449655(4t); .301 = .015779862t; t =
19.0768459 years
Determine how long it will take Natalie’s money to double under options 2 and 3.
Option 2:
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Option 3:
Solution: Option 2: 10.301 = 10.0013189741(12t); .301 = .0158276895t; t = 19.01920023 years
Option 3: 10.301 = 10.000042832412(365t); .301 = .0156338304t; t = 19.25503785 years
e. We also want to determine how long it would take Natalie’s money to double if it
was compounded continuously.
So we have 20,000 = 10,000e.036t  2 = e.036t.
Although we can employ the same method of writing each base as 10 to a power,
taking advantage of common logarithms, whenever we have a base of e, we
generally use natural logarithms instead. A natural logarithm is a logarithm with
a base of e. Instead of writing loge y = x  ex = y, we write ln y = x  ex = y.
Let’s explore natural logarithms. First, compute 2 times ½. What do you get?
What about 3/5 times 5/3? Why do we get the same answer?
Solution: In each case, we get 1. This is because the numbers being multiplied are inverses
under multiplication. Whenever we multiply two multiplicative inverses, we get the
multiplicative identity, 1.
Using your calculator, determine ln e. What do you get? Why?
Solution: We get 1 because natural logarithms are the inverse functions of the natural
exponential function.
So for our problem, we have 2 = e.036t. If we rewrite this equation as a logarithmic
equation, we get loge2 = .036t. We know that a logarithm with a base of e is a
natural logarithm, so we can rewrite again as ln 2 = .036t. Use your calculator to
help you finish solving the equation for t. (Your answer should be close to the
answers you got in part (d). If not, check that you are using your calculator
correctly.)
Solution: t = 19.255408835 years
f. Suppose Natalie invested her savings in an account that was compounded
continuously at an annual interest rate of 3.7%. How long would it take her
money to double?
First write the appropriate equation for this information. Then solve as in part (e).
Solution: y = 10,000e.037t; 2 = e.037t; ln 2 = .037t; t = 18.73370758 years
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Mathematics III
Unit 3
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Would your answer be different if she started with $20,000? What about
$100,000? What about $x? Explain. (Try some of the examples if you need to.)
Solution: No, the answer would not be different. If we are looking at when the money doubles,
each equation would reduce to the same equation, namely, 2 = e.037t.
g. The function f(x) = ex is a special case of the function f(x) = ax.
i. With that in mind, sketch a graph of f(x) = ex and list the domain, range,
intercepts, and asymptotes of f(x) = ex.
Solution:
Domain: all real numbers
Range: y > 0
Intercepts: (0, 1)
Asymptotes: y = 0
ii. Sketch the graph of Natalie’s earnings as a function of time, in years, if
she invests $1 into an account that is compounded continuously at an
annual interest rate of 3%. (First, write the function. The graph the
function.)
Solution: y = 1e.03t
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iii. How will the graph be different if Natalie invests $10,000? Will the
domain, range, intercept, and asymptote change? What, if anything,
changes? What kind of transformation is this?
Solution: If she invests $10,000 instead of $1, the domain, range, and asymptotes will remain
unchanged. However, because this is a vertical stretch by $10,000, all y-values will be 10,000
times their counterparts above. This results in a y-intercept of (0, 10,000).
iv. Suppose that Natalie invests $10,000. At the end of her mission, she
receives a $5,000 bonus. Write the equation for the amount of money that
Natalie has after the mission.
How will the graph of the equation be different from the graph of the
equation discussed in (iii)? What kind of transformation is this?
Solution: y = 10,000e.03t + 5,000; The graph of this equation will be a vertical translation
of the graph in (iii) up 5000 units. So the range, intercept, and asymptote will all be shifted
up 5,000 units. The new range is y > 5000. The intercept is (0, 15,000). The asymptote is y
= 5000.
6. During practice flights before the mission, the astronauts needed to work on maintaining
specific flying heights based on different amounts of air pressure.
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Mathematics III
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A pressure altimeter is an instrument that finds the height of an airplane above sea level
based on the pressure of the surrounding air. The height h and pressure p are related by
the equation h = -26,400 ln (p/2120) where h is measured in feet and p is measured in
pounds per square foot.5
a. The graph of f(x) = ln x is a special logarithm, as is g(x) =log10x. (Use your
graphing utility as a tool.) Sketch the graphs of f(x) and g(x). State the domain,
range, intercepts, and asymptotes of each.
Solution: Because the natural logarithm is a special logarithm, it will have characteristics
identical to other logarithms with a base greater than 1. (Characteristics are for both.)
Domain: x > 0; Range: all real numbers; Intercept: (1, 0); Asymptote: x > 0.
b. Does f(x) increase faster or slower than g(x)? What difference in the functions
accounts for this difference? Give an example of a function that increases faster
than both f(x) and g(x). Explain your reasoning.
Solution: Because the base in the natural log is smaller than the base in the common log, the
natural log graph increases faster. y = log2x increases faster than the natural and common log
graphs. (See The Population of Exponentia #5.)
c. Will the graph of the function stated in the problem be increasing/ decreasing?
Explain.
5
Based on a problem from Integrated Mathematics 3 by McDougal-Littell (2002).
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Solution: Because of the negative in front of the function, the graph will be reflected over the
x-axis. So instead of increasing, as does the standard natural logarithm, this graph will be
decreasing.
d. The function h = -26,400 ln (p/2120) is a transformation of the graph of f(x) = ln
x. Identify each transformation and explain how each would alter the graph of f(x)
= ln x.
Solution: The negative is a vertical reflection and makes the graph decreasing. It would also
change the intercept. The 1/2120 is a horizontal stretch. 26,400 is a vertical stretch and would
also change the intercept.
e. Consider the graph of h = -26,400 ln (p/2120).
If the intercept is usually (1, 0), what will the new intercept be? (Consider your
transformations. Which transformation impacts the x-values of your graph?)
Based on the transformations, what should be the domain, range, and asymptote?
Use this to help you set your graphing window.
Solution: The new intercept is (2120, 0). This is because the only transformation that impacted
the x-values was the horizontal stretch. This stretched the intercept from (1,0) to (2120,0).
The domain should not change; it will remain x > 0. The range should also remain
unchanged: all real numbers. The asymptote remains unchanged: x = 0.
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Mathematics III
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f. The control panel on the plane includes an instrument to measure height above
sea level and air pressure. Suppose the air pressure instrument malfunctioned.
Write an equation for determining the air pressure as a function of the height
above sea level.
We want to solve the above equation for p. First, isolate the natural logarithm
expression on one side of the equation.
Now use the fact that natural logarithms and natural exponentials, exponential
functions with a base of e, are inverses to rewrite the expression without
logarithms.
Solve for p.
Solution: h = -26400 ln(p/2120)  - h/26400 = ln(p/2120)  e^(- h/26400) = p/2120 
p = 2120e^(- h/26400)
g. If the plane’s height is 2500 ft, what is the surrounding air pressure? (Use your
equation in part (f).)
Solution: 1928.45 feet
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Mathematics III
Unit 3
2nd Edition
CULMINATING ACTIVITY: TRAVELING TO EXPONENTIA
This culminating task addresses all standards in the unit and is designed to serve as an
assessment task. Some parts of the items below are very explicit in order to assess student
mastery of particular topics. However, in general, the questions do not give the same level of
guidance as the learning tasks. In the learning tasks, students are guided quite explicitly to
solve problems and consider the meanings of the procedures. Such guidance is not included in
the assessment task in order to measure whether students have learned how to compare
graphs, use rational exponents, and understand exponential and logarithmic expressions and
functions As with the learning tasks, the process standards are embedded throughout the task.
A new solar system was discovered far from the Milky Way in 1999. One of the planets in the
system, Exponentia, has a number of unique characteristics. After much preparation, NASA sent
a group of astronauts to explore the planet.
1.
Motion sickness: The time finally came for the astronauts to travel to Exponentia. Before
leaving, the astronauts decided to take medication for motion sickness. Suppose that 85%
of this medication remains in the bloodstream after 1 hour. Three different doses are
taken by the astronauts. Natalie took 25 mg; Jason took 50 mg; and Derrick took 75 mg.
a. Write exponential equations for the amount of dimenhydrinate, motion sickness
medication, in each astronaut’s bloodstream after x hours.
Solution: Natalie: y = 25(.85)x; Jason: y = 50(.85)x; Derrick: y = 75(.85)x. Note that x is
number of hours.
b. How will the graphs of the three equations be similar? Different? Specifically
address domain, range, asymptotes, intercepts, and rate of increase/decrease.
Explain what accounts for any differences.
Solution: The graphs will have the same domain, range, asymptotes, and rate of decrease. The
domains are all real numbers. The ranges are y > 0. The asymptotes are y = 0. The rate of
decrease is 85%. Because the starting amounts are different, the three graphs will have
different y-intercepts. Natalie’s is (0, 25); Jason’s is (0, 50); and Derrick’s is (0, 75).
c. Express the equation representing Derrick’s remaining dimenhydrinate as a
function of Natalie’s equation. Then express Derrick’s remaining medication as a
function of Jason’s equation. What graphical transformation does this correspond
to?
Solution: Derrick’s equation is 3 times Natalie’s. If Natalie’s equation is called N(x), then
Derrick’s equation, D(x), can be expressed as follows: D(x) = 3N(x). Similarly, D(x) = 1.5J(x).
These are vertical stretches of N(x) and J(x).
d. Find the half-life for each concentration. That is, when does the amount in each
person’s bloodstream reach half of the original concentration? Explain your
solution.
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Solution: The answer will be the same for all three astronauts because the equations simplify
to the same equivalent equation: .5 = .85x. Using the method of rewriting exponential
equations as equations with both sides having base of 10 through the use of common
logarithms, we get 10-.301 = 10-.0706x. So x = 4.265 hours.
e. At what level might you consider each person’s bloodstream cleared of the
motion sickness medication? Why did you choose this level?
Solution: Answers will vary. One example: When there is less than 1 mg of the medication
remaining, I would consider it cleared. Technically, it will never reach 0 and this was pretty
close to 0.
f. Using your established criteria, find how long it takes for the dimenhydrinate to
clear each person’s bloodstream. Are the three answers the same? Why or why
not?
Solution: Answers will be based on criteria in (e). Answers based on above: Natalie: x = 19.81
hours; Jason: x = 24.07 hours; Derrick: x = 26.57 hours. Because each person started with a
different dosage, the time it takes for the remaining mg to be less than 1 mg will be greater for
those who took a higher dosage.
2.
Loudness of sound: While on the mission, the astronauts were exposed to a variety of
loud and soft sounds. The loudness of sound, D, measured in decibels (dB) is given by
the formula D = 10 log (I / 10-16), where I is the intensity measured in watts per square
cm (w/cm2). The denominator, 10-16, is the approximate intensity of the least sound
audible to the human ear.6
a. Describe how the graph of the loudness of sound, D, differs from the graph of f(x)
= log x. Specifically, name the transformations necessary for f(x) to be
transformed into D and state how each transformation alters the characteristics of
the logarithm graph (domain, range, asymptote, intercept, increasing/decreasing).
Graph D.
Solution: The 10 is a vertical stretch and will change all the y-values by a scale factor of 10. It
does not change the domain, range, asymptote, or intercept, but it does make the graph
increase faster. 1016 is a horizontal shrink by a factor of 1016, so the intercept will now be
(1016, 0).
(Note: The graphing program being used for this teachers’ edition can not handle the
numbers needed for this graph. Student graphs should have an x-intercept of (1016, 0).
Students should substitute in some reasonable values, e.g. 1013, 1011, etc., to help them obtain
an accurate graph.)
6
Problems 2 and 3 are adapted from Discovering Advanced Algebra: An Investigative Approach from Key
Curriculum Press, 2004.
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Mathematics III
Unit 3
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b. If a normal conversation is held at an intensity of 3.16 x 10-10 w/cm2, how many
decibels is this? Show your work. Simplify as much as possible before using the
calculator. (You should use some properties of exponents.)
Solution: D = 10 log (I / 10-16) = 10 log (3.16 x 10-10 / 10-16) = 10 log (3.16 x 106)  65 dB
c. Suppose the whisper of the ventilation system in the space shuttle had an intensity
of 10-15 w/cm2. How many decibels is this? Do not use a calculator. Explain how
you determined the answer.
Solution: D = 10 log (I / 10-16) = 10 log (10) = 10 dB. We know that log (10) means the
exponent, a, that the base 10 is raised to in order for 10a = 10. So a = 1. Therefore, the
equation is now 10 (1) = 10 dB.
d. The loudest a rock concert may be held is 120 dB. This is also how loud a space
shuttle launch is from a viewing area for non-essential NASA personnel. What is
the intensity of the launch from this site? Leave your answer in exponential
notation. (You will need to use that logarithms and exponential functions are
inverses.) 7
Solution: D = 10 log (I / 10-16)  12 = log (I/10-16)  1012 = I/10-16  10-4 w/cm2 = I
e. Closer to the launch pad, the sound reaches decibels levels as loud as 180 to 200
dB. What is the range of sound intensity at this site? Leave your answer in
exponential notation.
Solution: 180 < 10 log(I / 10-16) < 200  18 < log(I / 10-16) < 20  102 < I < 104 w/cm2
f. How many more times more intensity is the launch at the closer observation site
than at the farther site in part (c)? Express your answer in exponential notation.
Solution: Between 106 and 108 times the w/cm2.
3. Intensity of light: The intensity, I, of a beam of light after passing through t cm of liquid
is given by I = Ae-kt, where A is the intensity of the light when it enters the liquid and k is
a constant for the particular liquid.
During the planetary rover’s exploration of Exponentia, it encountered a number of
different liquid like substances. The NASA scientists would like the astronauts to run
some tests on these liquids.
a. To make sure that they understood how to use the formula, the scientists asked the
astronauts the following questions: If the light intensity in a lake is measured at a
7
Information obtained from the Space Shuttle Recording Project.
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Mathematics III
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depth of 50 cm, and the light intensity was 80% of the light intensity at the
surface, what is the constant k for the lake water?
Solution: .8 = 1e-k(50)  loge.8 = -50k  ln .8 = -50k  k = .0045
b. If the light intensity is measured as 1% of the intensity at the surface of the lake,
at what depth was the reading taken?
Solution: .01 = 1e-.0045(t)  loge.01 = -.0045t  ln .01 = -.0045t  t = 1023.27 cm (or
10.2327 m)
c. Suppose that the intensity of light measured 1 m under the surface of the liquid
found on Exponentia is 35% of the intensity of light measured at the surface.
What is the constant k?
Solution: .35 = e-k(100)  ln .35 = -100k  k = .0105
d. At what depth would you need to measure the lake and the substance on
Exponentia for the light intensity to be half of what it is at the surface? Use your
answers to determine which allows light through more easily.
Solution: Lake: .5 = e-.0045t  ln .5 = -.0045t  154.03 cm or 1.54 m
Exponentia: .5 = e-.0105t  ln .5 = -.0105t  66.01 cm
Because it takes less depth for the light to reach 50% of its intensity on Exponentia, this
substance is more resistant to allowing light through. So the lake water allows light
through more easily.
e. Describe the transformations of the graph of the light intensity in the lake from
the graph of y = ex. How will they be similar or different?
Solution: The equation for light intensity in the lake is I = e-.0045t. The negative sign
reflects the graph of y = ex over the y-axis, so it is now decreasing. The .0045 is a
horizontal stretch by a factor of 1/.0045 = 2000/9.This makes the decrease much slower
than the original increase. The domain, range, intercept, and asymptote all remain
unchanged.
f. How will the graphs of the light intensity for the lake water and light intensity for
the substance on Exponentia compare?
Solution: Because the horizontal stretch is only 1/.0105 = 2000/21  95.24 in the
Exponentia graph but it is 2000/9  222.2 in the lake graph, the Exponentia graph will not
take as long to decrease. All other characteristics will be the same.
g. Consider the equations of the light intensity in the lake water and in the substance
discovered on Exponentia. Write one equation as a transformation of the other.
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Mathematics III
Unit 3
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Solution: If the lake equation is L(x) = e-.0045x and the Exponentia equation is E(x) = e.0105x, we can write L(x) = E((7/3)x) or E(x) = L((3/7)x).
4. As part of an ongoing space project, the astronauts grew a number of plants while on
board the space shuttle. They were charged with keeping up with records on the plants’
growth.
Suppose that the plant was measured at noon each day, with the following results.
Day
Height (cm)
0
2.5
1
5
2
10
3
20
a. Write an exponential equation to model the growth pattern. (Hint: By how much, in
relation to the previous day, is the height changing each day?) Specify what each
variable means and the units of measurement for each variable.
Solution: y = 2.5(2x), where x is the day at noon and y is the height in centimeters.
b. Suppose that the astronauts forgot to measure the plant on the 5th day. Assuming the
rate of growth remained consistent, what was the height of the plant on day 5 at
noon?
Solution: y = 2.5(25) = 80 cm
c. On the 8th day of the mission, the astronauts, following orders, removed part of the
plant from the shuttle and placed it on Exponentia so that they could determine how
the plant grows in the new atmosphere. However, the astronauts forgot to measure the
height before removing it from the shuttle. If they removed it at 6 pm on the 8th day,
what should the height have been?
Solution: y = 2.5(28.25) = 761.09 cm or 7.6109 m
d. After being placed in Exponentia’s atmosphere, the plant continued to grow. The
following table provides the data for the plant after removal from the shuttle. Write an
exponential equation to model the growth pattern.
Days (on
Exponentia, at
6pm)
Height (cm)
1
2
3
4
1141.5
1712.25
2568.375
3852.5625
Solution: y = 761(1.5)x
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Unit 3: Page 63 of 66
Mathematics III
Unit 3
2nd Edition
e. Explain how the graphs of the equations from (a) and (d) will compare. Specially
address their domains, ranges, asymptotes, intercepts, and rate of change.
Solution: Because the starting values are different, the y-intercepts will be different. At time 0
for the original plant, it was 2.5 cm tall. At time 0 for the removed plant, it was 761 cm tall. So
the first intercept is (0, 2.5) and the second is (0. 761). Also the rates of change will be
different. The plant on the shuttle grows faster and so will increase faster than the plant on
Exponentia. Otherwise, the domain, range, and asymptotes remain the same.
f. How long will it take the plant that remained on the shuttle to triple its height? How
long, from the time it was removed from the shuttle, will it take the plant on
Exponentia to triple in height?
Solution: Shuttle: 7.5 = 2.5(2x)  3 = 2x  log (3) = log (2) * x  1.58 days
Exponentia: 2283 = 761(1.5x)  3 = 1.5x  log (3) = log (1.5)  2.71 days
Explain why there is / is not a difference in your two answers. If they are different, give a
scenario for which they would have the same tripling time.
Solution: There is a difference because the rates are different. If they had the same rate, say 2,
then their tripling times would be the same.
g. Assuming that the plants survive, how long will it take each one to reach 0.5 km tall?
What time will this be?
Solution: Shuttle: 50,000 cm = 2.5(2x)  20,000 = 2x  x = 14.29 days. This would be at
approximately 6:54 pm on the 14th day of the mission.
Exponentia: 50,000 cm = 761(1.5x)  65.7 = 1.5x  x = 10.32 days. This would be at
approximately 1:43 am on the night of the 10th day that the plant was on Exponentia (10.32
days translates to 7 hours, 43 minutes after the 10th complete day on the planet. Since the plant
was placed on the planet at 6 pm, 7 hours later makes the time 1:00am.)
5. Orbital time and orbital radius: German astronomer Johannes Kepler discovered in
1619 that the square of a planet’s orbital period is proportional to the cube of its orbital
semi-major axis. This is known as Kepler’s 3rd Law of Motion.
Another way to state this law is that the mean orbital radius, measured in astronomical
units (AU), is equal to the time of one complete orbit around the sun, measured in years,
raised to the power 2/3.8 (One AU is approx. 1.5 x 108 km or 9.3 x 107 miles.)
8
Problem 5 is adapted from Integrated Mathematics 3 by McDougal-Littell, 2002.
Georgia Department of Education
Kathy Cox, State Superintendent of Schools
June, 2010
Copyright 2009 © All Rights Reserved
Unit 3: Page 64 of 66
Mathematics III
Unit 3
2nd Edition
a. Write the equation that relates mean orbital radius, r, to the orbital time, t.
Solution: r = t2/3
b. If Neptune requires 165.02 Earth years to orbit the sun, what is Neptune’s orbital
radius? How many km is this? How many miles is this?
Solution: 30.09 AU = 4,512,830,074 km = 27,979,554,646 miles
c. Saturn has an orbital radius of 9.542 AU. How many Earth years does it take for
Saturn to orbit the sun?
Solution: 9.542 = t2/3  9.5423/2 = t  t = 29.47536 years
d. Complete the chart below.
Planet
Orbital
Radius
(AU)
Mercury Venus
Earth
Mars
Jupiter
.387
.615
Orbital
Time
(years)
1.8795
Saturn
Uranus
9.542
19.181
11.861
Neptune
165.02
Solution:
Planet
Mercury Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Orbital
Radius
(AU)
.387
.7232
1
1.523
5.201
9.542
19.181
30.086
Orbital
Time
(years)
.2408
.615
1
1.8795
11.861
29.475
84.008
165.02
e. After observing Exponentia and its sun, Napier, for a number of years, the NASA
astronomers determined that, even though the radius and time are still
proportional, the exponent in the above equation would be different for their solar
system. The astronomers determined that if Saturn was in Exponentia’s solar
system, it would take 70 Earth years to orbit the sun.
Georgia Department of Education
Kathy Cox, State Superintendent of Schools
June, 2010
Copyright 2009 © All Rights Reserved
Unit 3: Page 65 of 66
Mathematics III
Unit 3
2nd Edition
Use the orbital radius above and determine the equation for the relationship
between a planets orbital radius and time in Exponentia’s solar system.
Solution: 9.542 = (70)x  .5309416578  0.53; r = t.53
f. One of the NASA astronomers, Delia, proposed using the following equation for
the relationship between orbital radius and time in Exponentia’s solar system: t
=(tm)5/4, where tm is the orbital time if the planet was in our solar system, the
Milky Way. Evaluate Delia’s proposed equation. How does this relate to your
equation in (e)?
Solution: Delia’s equation is equivalent to t = (r3/2)5/4 = r15/8. This is the inverse of the equation
in (e). (e) could be written as t = r1/.53 = r1.89. 15/8 = 1.875, so the exponent values are quite
close.
6. Design a Viètian (vee-et-ee-an). The inhabitants of Exponentia are Vietians. Use the graphs
of exponential and logarithmic functions to design these creatures encountered by the
astronauts on their visit to Exponentia. Your design must meet the following requirements:
The design must use the basic exponential and logarithmic graphs as well as the natural
exponential, e, and the natural logarithm. Use combinations of vertical and horizontal shifts,
vertical and horizontal stretches or shrinks, and reflections through the x-axis or y-axis of the
basic functions. Horizontal and vertical lines, as well as any other functions learned in previous
courses and units, may also be used.
For each equation used, label it in your picture (numbered). On a supplemental page, list the
equations (with their numbers), the domains, ranges, asymptotes, and intercepts. Be sure to show
any associated work.
Solution: Answers will vary. Make sure that all requirements are met and are mathematically
correct. Alter this activity to fit your students. An extension of this activity is to have the
students write a program in their graphing calculators that will draw their picture.
Another extension: Students could also animate their picture to simulate waving, walking, etc.
Georgia Department of Education
Kathy Cox, State Superintendent of Schools
June, 2010
Copyright 2009 © All Rights Reserved
Unit 3: Page 66 of 66