Download Using Graphs and Properties to Solve Equations with Exponents

LESSON
13.1
Name
Using Graphs and
Properties to Solve
Equations with
Exponents
A1.9.D …graph exponential functions that model growth...in...real-world problems.
Also A1.9.B, A1.9.C
In previous lessons, variables have been raised to rational exponents and you have seen how to simplify and solve
equations containing these expressions. How do you solve an equation with a rational number raised to a variable? In
x
certain cases, this is not a difficult task. If 2 x = 4 it is easy to see that x = 2 since 2 2 = 4. In other cases, like 3(2) =
96, where would you begin? Let’s find out.
Graph exponential functions that model growth and decay and identify
key features, including y-intercept and asymptote, in mathematical and
real-world problems. Also A1.9.B, A1.9.C


Mathematical Processes
Solve for 3(2) = 96 for x.
x
Let ƒ(x) = 3(2) . Complete the table
for ƒ(x).
x
A1.1.C
Select tools, including real objects, manipulatives, paper and pencil, and
technology as appropriate, and techniques, including mental math,
estimation, and number sense as appropriate, to solve problems.
x
f(x)
1
6
2
3
4
Language Objective
5
6
1.B, 3.E, 3.F, 3.H
When the bases are equal, use the Equality of Bases
Property. When the bases are not equal, graph each
side of the equation as its own function and find the
intersection.
Solving Exponential Equations Graphically
Explore 1
A1.9.D
7
© Houghton Mifflin Harcourt Publishing Company
Essential Question: How can you solve
equations involving variable
exponents?
Resource
Locker
Essential Question: How can you solve equations involving variable exponents?
The student is expected to:
ENGAGE
Date
13.1 Using Graphs and Properties
to Solve Equations with
Exponents
Texas Math Standards
Explain to a partner how to use a graph to find the solution to an
equation with a variable exponent.
Class

g(x)
1
96
2
6
96
96
96
96
96
7
96
4
5

12
24
48
96
192
384
x
The graphs intersect at point(s):
Module 13
(5, 96)
96
72
(4, 48) f(x)
48
(3, 24)
24
(1, 6)
0
-2
Let g(x) = 96. Complete the table
for g(x).
3
Using the table of values, graph ƒ(x)
on the axes provided.
y

(2, 12)
x
1 2 3 4 5 6 7 8
Using the table, graph g(x) on the
same axes as ƒ(x).
y

96
72
g(x)
(5 , 96)
f(x)
48
24
x
0
-2
1 2 3 4 5 6 7 8
(5, 96) This means that ƒ(x) = g(x) when x = 5 .
ges
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DO NOT Key=TX-A
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Lesson 1
605
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A1_MTXESE353879_U5M13L1 605
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A1.9.D …grap
A1.9.C
Also A1.9.B,
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96 for
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 Solve
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 Let
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Using the
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y
g(x)
x

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x
24
(2, 12)
(1, 6)
8
5 6 7
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96. Comp
Let g(x) =
for g(x).
(4, 48) f(x)
48
-2


A1_MTXE
Lesson 13.1
Using the
provided.
on the axes
y
(5, 96)
96
6
1
Turn to these pages to
find this lesson in the
hardcover student
edition.
table of
f(x)
x
Module 13
605
ally
fy and solve In
l
how to simpli to a variable?
x
Exponentia ents and you have seennumbe
r raised
=
like 3(2)
1 Solving
rational
rational expon
other cases,
Explore
on with a
raised to
22 = 4. In
an equati
have been
x = 2 since
do you solve
s, variables
sions. Howx 4 it is easy to see that
us lesson
=
In previo
these expres
task. If 2
containing
a difficult
equations
this is not
find out.
certain cases,
begin? Let’s
would you
ƒ(x)
96, where
values, graph
x
View the Engage section online. Discuss why a town
government might need to know the rate at which the
town’s population is growing. Then preview the
Lesson Performance Task.
HARDCOVER PAGES 449456
Resource
Locker
ents?
le expon
ing variab
involv
equations
This means
that ƒ(x)
7 8
4 5 6
1 2 3
when x =
= g(x)
5 .
Lesson 1
605
20/02/14
1:10 AM
20/02/14 1:10 AM
Reflect
EXPLORE 1
Discussion Consider the function h(x)= -96. Where do ƒ(x) and h(x) intersect?
The graphs would not intersect as f(x) is always greater than 0. Raising any positive
1.
Solving Exponential Equations
Graphically
number to a positive exponent yields a positive number.
Divide both sides of the equation 3( 2 = 96 by 3 (an algebraic step) and utilize the same method as
)x
2.
in Explore 1 to graph each side of the equation as a function. The point of intersection would be:
(5, 32)
.
Is this the same point of intersection? Is this the same answer? Can this be done? Elaborate as to why or why
not.
INTEGRATE TECHNOLOGY
It is not the same point of intersection. The y-values of the points are different. They
Students can use graphing calculators to solve
an exponential equation by the method shown
in the Explore activity. Students should enter the
appropriate exponential function and constant
function, graph both functions, and use the
calculator’s intersect feature to find their point
of intersection.
do represent the same solution because the equations are equivalent by the Division
Property of Equality.
Solving Exponential Equations Algebraically
Explore 2
Recall the example 2 = 4, with the solution x = 2. What about a slightly more complicated equation? Can an
x
equation like 5 (2) = 160 be solved using algebra?
x
A
Solve 5 (2) = 160 for x. The first step in isolating the term containing the variable
x
on one side of the equation is to divide each side of the equation by 5 .
QUESTIONING STRATEGIES
5( 2 ) x _
_
= 160
5
B
When solving an exponential equation of the
form ab x = c graphically, what two functions
do you graph? the exponential function f(x) = ab x
and the horizontal line g(x) = c
5
Simplify.
C
(2) = 32
x
Rewrite the right hand side
as a power of 2.
5
x
(2) = (2)
D
Solve.
x= 5
3.
Discussion The last step of the solution process seems to imply that if b x = b y then x = y. Is this true for
all values of b? Justify your answer.
No, it is not true. For example, 0 5 = 0 8 but 5 ≠ 8, or 1 7 = 1 958 but 7 ≠ 958.
4.
In Reflect 2, we started to solve 3(2) = 96 algebraically. Finish solving for x.
x
3(2) = 96
x
2 x = 32
2x = 25
How does graphing these two functions on the
same grid help you determine the value of the
exponent? The value of the exponent is the x-value
of the point of intersection.
© Houghton Mifflin Harcourt Publishing Company
Reflect
EXPLORE 2
Solving Exponential Equations
Algebraically
x=5
QUESTIONING STRATEGIES
Module 13
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Lesson 1
PROFESSIONAL DEVELOPMENT
A1_MTXESE353879_U5M13L1.indd 606
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Learning Progressions
In this lesson, students continue to build on their understanding of geometric
sequences and exponential functions. They learn the Equality of Bases Property,
which states that If b > 0 and b ≠ 1, then b x = b y if and only if x = y. They learn
to solve equations involving variable exponents either by using the Equality of
Bases Property or by graphing. They also begin to model real-world situations
using exponential equations, which can then be solved by either method. Work
with exponential functions will continue as students learn about exponential
growth and decay models and exponential regression.
In the equation 4 x = 64, how can you
evaluate x? Write 64 as a power of 4.
3
64 = 4 , so x = 3.
Assuming that x is an integer in the equation
b x = c, what must be true for this method to
work? The value of c must be a power of b.
Using Graphs and Properties to Solve Equations with Exponents
606
Solving Equations by Equating Exponents
Explain 1
EXPLAIN 1
Solving the previous exponential equation for x used the idea that if 2 x = 2 5, then x = 5. This will be a powerful tool
for solving exponential equations if it can be generalized to if b x = b y then x = y. However, there are values for which
this is clearly not true. For example, 0 7 = 0 3 but 7 ≠ 3. If the values of b are restricted, we get the following property.
Solving Equations by Equating
Exponents
Equality of Bases Property
Two powers with the same positive base other than 1 are equal if and only if the exponents are equal.
Algebraically, if b > 0 and, b ≠ 1, then b x = b y if and only if x = y.
QUESTIONING STRATEGIES
x


5·
Multiply both sides by _
2
Simplify.
Rewrite the right side as a power of 5.
x = 4
Equality of Bases Property.
( )
5 x=_
250
2 _
27
3
( )
5 x
250
_
2 _
3
27
_=_
2
Divide both sides by 2 .
2
(_35 )
© Houghton Mifflin Harcourt Publishing Company
Discuss with students the limitations on the Equality
of Bases Property. Have students give examples to
show why the property does not apply when the base
is 0, 1, or –1. For example: 05 = 0 8 = 0 but 5 ≠ 8;
4
6
1 20 = 1 99 = 1 but 20 ≠ 99; and (–1) = (–1) = 1
but 4 ≠ 6.
2 ( 5 ) x = 250
_
5
5·_
5
2 (5) x = 250 · _
_
2 5
2
5 x = 625
5x = 54
How does this step compare to isolating a
variable on one side of a linear equation? It is
done for the same reason. By isolating the power,
you have isolated the variable as well. Then you can
compare the exponents in the final equivalent
expression.
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Reasoning
Solve by equating exponents and using the Equality of Bases Property.
Example 1
In an equation such as 36(2) = 576, what
property of equality can you use to isolate 2 x?
Explain how. Division Property of Equality; divide
both sides by 36.
(_35 )
x
x
125
=_
27
=
(_53)
Simplify.
3
x= 3
Rewrite the right side as a power of
5
_
.
3
Equality of Bases Property.
Reflect
5.
Suppose while solving an equation algebraically you are confronted with:
5 x = 15
5x = 5
Can you find x using the method in the examples above?
No, you cannot. It is not possible because 15 is not a whole number power of 5.
AVOID COMMON ERRORS
Some students may misread the base b in an
expression b x as a coefficient of x and try to divide
both sides of the equation by b to isolate the variable.
Remind them that when a number is raised to a
power, it cannot be treated as a single factor. They
must use the properties of equality to isolate b x,
then use the Equality of Bases Property to solve for
the variable.
607
Lesson 13.1
Module 13
607
Lesson 1
COLLABORATIVE LEARNING
A1_MTXESE353879_U5M13L1 607
Peer-to-Peer Activity
Have students work in pairs. Have each student write an equation involving a
variable exponent in the form b x = c. After students exchange equations, each
partner should first decide whether c can be expressed as a whole number power
of b. If so, the student should rewrite c as a power of b and solve for x. If not, the
student should use a graphing calculator to graph each side of the equation as a
separate function and use the intersect feature to find the x-coordinate of the
intersection point, which is the solution to the original equation. Have students
check each other’s work.
20/02/14 1:10 AM
Your Turn
EXPLAIN 2
Solve by equating exponents and using the Equality of Bases Property.
6.
2 (3) x = 18
_
3
7.
_3 (_4 ) = _8
_2 (3) = 18
x
3
2 3
3
x
x
2
3
Solving a Real-World Exponential
Equation by Graphing
x
_2 ⋅ _3 (_4 )
_3 ⋅ _2 (3) = 18 ∙ _3
2 3
3 x = 27
()
x
8
3 _
4 =_
_
2 3
3
2 3
=
16
(_43 ) = _
9
4
4
_
_
(3) = (3)
_2 ∙ _8
3
3
QUESTIONING STRATEGIES
x
3x = 33
x
x=3
If a population grows by 5% each year, by
what factor is the population multiplied each
year? Explain. 1.05; if the population is p one year, it
will be p + 0.05p = 1.05p the next year.
2
x= 2
Explain 2
Solving a Real-World Exponential
Equation by Graphing
Why is it appropriate to round a prediction
involving time to the nearest year?
A prediction is usually just an estimate, so rounding
is appropriate.
Some equations cannot be solved using the method in the previous example
because it isn’t possible to write both sides of the equation as a whole number
power of the same base. Instead, you can consider the expressions on either
side of the equation as the rules for two different functions. You can then
solve the original equation in one variable by graphing the two functions. The
solution is the input value for the point where the two graphs intersect.
Solve by graphing two functions.
An animal reserve has 20,000 elk. The population is increasing at a rate of
8% per year. There is concern that food will be scarce when the population
has doubled. How long will it take for the population to reach 40,000?
Analyze Information
Identify the important information.
• The starting population is
• The ending population is
• The growth rate is
20,000.
40,000
.
8% or 0.08 .
Formulate a Plan
With the given situation and data there is enough information to write and solve an exponential
model of the population as a function of time. Write the exponential equation and then solve it using
a graphing calculator.
Set ƒ(x) = the target population and g(x) = the exponential model .
Input Y 1 = ƒ(x) and Y 2 = g(x) into a graphing calculator, graph the functions,
and find their intersection .
Module 13
608
© Houghton Mifflin Harcourt Publishing Company · Image Credits: © James
Prout/Alamy
Example 2
Lesson 1
DIFFERENTIATE INSTRUCTION
A1_MTXESE353879_U5M13L1.indd 608
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Graphic Organizers
Have students complete a graphic organizer that shows when to solve an equation
involving a variable exponent algebraically and when to solve it graphically.
Solving b x = c (where b > 0, b ≠ 1, and c > 0)
Value of c
Solution Method
d
c = b for some
Algebraic:
whole number d.
c is not a power of b.
b x = b d, so x = d.
Graph: intersection of ƒ(x) = b x and g(x) = c
Using Graphs and Properties to Solve Equations with Exponents
608
Solve
INTEGRATE TECHNOLOGY
Write a function P(t) = ab t, where P(t) is the population and t is the number of years since the
population was initially measured.
When solving exponential equations
graphically, have a student demonstrate how to
identify the two functions to be graphed, enter them
into a graphing calculator, and find the solution by
finding the point of intersection. Discuss how to
adjust the viewing window so that the graph and the
point of intersection are clearly visible.
a represents the initial population of elk
a = 20,000
b represents the yearly growth rate of the elk population
b = 1.08
( 1.08 ) .
t
The function is P(t) =
20,000
()
To find the time when the population is 40,000, set the function or P t
equal to 40,000 and solve for t .
( 1.08 ) .
t
40,000 = 20,000
Write functions for the expressions on either side of the equation.
40,000
ƒ(x) =
)
(
g(x) = 20,000 1.08
x
Using a graphing calculator, set Y 1 = ƒ(x) and Y 2 = g(x). View the graph.
Use the intersect feature on the CALC menu to find the intersection of the two graphs.
The approximate x-value where the graphs intersect is 9.006468 .
Therefore, the population will double in just a little over 9 years.
© Houghton Mifflin Harcourt Publishing Company
Justify and Evaluate
Check the solution by evaluating the function at t = 9 .
P
( 9 ) = 20,000 ⋅ (1.08)
= 20,000 ⋅ ( 1.9990 )
9
= 39,980
Since 39,980 ≈ 40,000, it is
in 9 years.
accurate to say the population will double
This prediction is reasonable because 1.08
Module 13
9
609
≈ 2 .
Lesson 1
LANGUAGE SUPPORT
A1_MTXESE353879_U5M13L1.indd 609
Connect Context
Support students in interpreting the language used in problem statements. Explain
that the word suppose at the beginning of a problem signals that what follows is a
hypothetical example, meaning that readers should use their imaginations to
consider a possible scenario.
Often, a problem will be followed by the question, Why or why not? Explain that
the question is phrased this way so as not to give away the answer. Students should
understand that they need to explain either why a result is true or why it is not
true, depending on the situation.
609
Lesson 13.1
13/11/14 4:24 PM
Your Turn
ELABORATE
Solve using a graphing calculator.
8.
9.
There are 225 wolves in a state park. The population is increasing at the rate of 15% per year. You want to
make a prediction for how long it will take the population to reach 500.
Y 1 = 500
Graph
The intersection point is (5.713341, 500). The wolf
x
Y 2 = 225(1.15)
population will reach 500 in approximately 5.7 years.
There are 175 deer in a state park. The population is increasing at the rate of 12% per year. You want to
make a prediction for how long it will take the population to reach 300.
Y 1 = 300
The intersection point is (4.756046, 300). The deer
Graph
x
Y 2 = 175(1.12)
population will reach 300 in approximately 4.8 years.
QUESTIONING STRATEGIES
What is the shape of the graph of an
exponential function of the form f(x) = b x
when b > 1? It is a curve that rises in greater and
greater amounts as x increases.
What is the shape of the graph of a function
ƒ(x)= b x when 0 < b < 1? It is a curve that
falls more and more gradually as x increases.
Elaborate
10. Explain how you would solve 0.25 = 0.5 x. Which method can always be used to solve an exponential
equation?
Possible answer: Algebraically.
0.25 = 0.5 x
(0.5)2 = (0.5) → x = 2
SUMMARIZE THE LESSON
Exponential equations can always be solved graphically.
How can you solve an equation where the
variable is an exponent? First, use the
properties of equality to isolate the number raised
to a variable power. Then check whether the
constant on the other side of the equation can be
written as a whole-number power of the same base.
If it can, use the Equality of Bases Property to solve.
If not, graph each side of the equation as its own
function and find the x-value of the point of
intersection.
x
x
11. What would you do first to solve the equation __14 (6) = 54?
Multiply each side of the equation by 4 to isolate the power.
12. How does isolating the power in an exponential equation like __14 (6) = 54 compare to isolating the variable
in a linear equation?
Both are done for the same reason. By isolating the power, you have isolated the variable
x
as well. Then you can compare the exponents in the final equivalent expressions.
situation, 2 = 0.99 x, has no solution for x > 0. The graphing calculator will show a
horizontal line at 2 and an exponential function with a y-intercept of 1 decreasing towards
the positive x-axis.
14. Solve 0.5 = 1.01 x graphically. Suppose this equation models the point where a population increasing at a
rate of 1% per year is halved. When will the population be halved?
Since x = -69.66072, you would have to go back in time, which is not possible. Seventy
years or so ago the population was half of what it is now.
© Houghton Mifflin Harcourt Publishing Company
13. Given a population decreasing by 1% per year, when will the population double? What will this type of
situation look like when graphed on a calculator?
It will never double as the population is decreasing. The equation representing this
15. Essential Question Check-In How can you solve equations involving variable exponents?
When the bases are equal, use the Equality of Bases Property. When there are not equal
bases on both sides of the equation, graph each side of the equation as its own function
and find the intersection.
Module 13
A1_MTXESE353879_U5M13L1 610
610
Lesson 1
20/02/14 1:10 AM
Using Graphs and Properties to Solve Equations with Exponents
610
Evaluate: Homework and Practice
EVALUATE
1.
• Online Homework
• Hints and Help
• Extra Practice
Would it have been easier to find the solution to the equation in Explore 1,
x
3(2) = 96, algebraically? Justify your answer. In general, if you can solve an
exponential equation graphing by hand, why can you solve it algebraically?
Yes, 3(2) = 96 becomes (2) = 32 after dividing both sides of the
equation by 3 and 32 is an integer power of 2.
x
x
In general, the input-output tables for f(x) and g(x) have integers in the
domain and the values in the range are easy to calculate.
ASSIGNMENT GUIDE
Concepts and Skills
Practice
Explore 1
Solving Exponential Equations
Graphically
Exercises 2–3
Explore 2
Solving Exponential Equations
Algebraically
Exercises 1, 3
Example 1
Solving Equations by Equating
Exponents
Exercises 4–16,
24 –25
Example 2
Solving a Real-World
Exponential Equation by
Graphing
Exercises 17–23
2.
The equation 2 = (1.01) models a population that has doubled. What is the rate of increase? What does x
represent?
x
The rate of increase is 1% per unit time. x is number of units of time.
3.
Can we solve equations using both algebraic and graphical methods?
Yes. We can simplify the equation algebraically and then use graphing.
Solve the given equation.
x
4. 4(2) = 64
5.
4(2)
64
____
= __
4
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Communication
Circulate as students solve the practice exercises.
Invite students to explain their reasoning as they
begin a new problem.
© Houghton Mifflin Harcourt Publishing Company
7.
7(3) = 63
x
7
3x = 9
2x = 24
3x = 32
x=4
x=2
75
(_14 )(_56 ) = _
432
x
8.
5
75
1 _
= 4 ⋅ ___
4⋅ _
4 6
432
x
x
x
2
x=2
()
Exercise
611
Lesson 13.1
6 x = 216
6x = 63
x=3
x
49
7 =_
2_
2
2
9.
2( 2 )
2
____
= __
_7
2
_7
2
_7
2
x
2
49
__
= 4
x
7
= _
2
3(11) = 3993
x
3(11)
3993
______
= ____
x
49
__
3
3
()
() ()
11 x = 1331
x
11 x = 11 3
2
x=3
x=2
Module 13
A1_MTXESE353879_U5M13L1 611
x
7
2 x = 16
6 x = 54
_
4
6
= 4 · 54
4 · __
4
x
4
( )( )
25
(_56) = __
36
5
(_6) = (_56)
6.
7(3)
63
____
= __
x
Lesson 1
611
Depth of Knowledge (D.O.K.)
Mathematical Processes
1
2 Skills/Concepts
1.D Multiple representations
2
2 Skills/Concepts
1.A Everyday life
3
2 Skills/Concepts
1.D Multiple representations
4–12
1 Recall
1.F Analyze relationships
13–16
2 Skills/Concepts
1.F Analyze relationships
17–22
2 Skills/Concepts
1.A Everyday life
20/02/14 1:10 AM
()
x
1
2 _
9
9x = 92
9
81
9
9
(_21 )(_32 ) = (_41 )(_1627 )
( )( ) ( )( )
8
(_23) = __
27
(_23) = (_23)
16
1 _
2
1 __
2 _
=2 _
4 27
2 3
x
x
x
3
x=3
x=2
()
x
( )
15.
16
___
8
8
13
13
When solving an equation involving a variable
exponent, suggest that students try to structure the
solution so that they are solving an equation of the
form b x = c y. If c = b, then x = y; if c ≠ b, then they
should solve by graphing.
8
(_25 )(_25 ) = _
125
x
x
27
x
x
3
(_2) = (_2)
3
169
x
x
(_2) = __8
3
13
8
(_52)(_25)(_25) = (_52)___
125
4
(_25) = __
25
(_25) = (_25)
8(3)
4( 27 )
____
= ____
x
2
x=2
16
2 = (4) _
14. (8) _
27
3
_2
2
x
2
(__4 ) = (__4 )
x
2
(_1) = (_1)
x
x
x
16
(__4 ) = ___
x
(_1) = __1
x=2
16.
2
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Patterns
32
___
( ) ___
____
= 169
4
2 __
13
x
2
x
32
4 =_
12. 2 _
13
169
2
__
( ) __
____
= 81
9 x = 81
13.
( )
x
1 =_
2
11. 2 _
81
9
10. 2(9) = 162
3
2
x=2
x=3
8 _
8
(_25 ) (_52 ) = (_
125 )( 125 )
x
2x
2x
2x
( )
2
= ((_
5) )
2
= (_
5)
8
= ___
125
2
3 2
6
© Houghton Mifflin Harcourt Publishing Company · Image Credits: ©prudkov/
Shutterstock
(_25)
(_25)
(_25)
x
2x = 6
6
2x
__
=_
2
2
x=3
17. There is a draught and the oak tree population is decreasing at
the rate of 7% per year. If the population continues to decrease
at the same rate, how long will it take for the population to be
half of what it is?
The model for the oak tree population is P(t) = P i(0.93) ,
t
where t is the time in years, P i is the initial population,
and P(t) is the population in year t. To find when the
i
population is half of its initial value, solve P(t) = __
for t.
2
P
Pi
t
__
= P i(0.93)
2
P
__
P (0.93)
__2 = ______
i
t
i
Pi
Pi
_1 = (0.93) t
2
t ≈ 9.55
Using the calculator solution method
from Explain 2, the population will
reach half of its original value in
approximately 9.6 years.
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Exercise
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Lesson 1
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Depth of Knowledge (D.O.K.)
Mathematical Processes
23
3 Strategic Thinking
1.A Everyday life
24–25
3 Strategic Thinking
1.G Explain and justify arguments
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612
18. An animal reserve has 40,000 elk. The population is increasing at a rate of 11% per
year. How long will it take for the population to reach 80,000?
AVOID COMMON ERRORS
The model for population is P(t) = 40,000(1.11) , where t is the time in
t
Students may be confused by complicated
equations that involve variable exponents as well as
additional factors. Remind them to first apply the
properties of equality to isolate the number with the
variable exponent, then use the Equality of Bases
Property to solve.
years and P(t) is the population in year t. To find when the population is
80,000, solve P(t) = 80,000 for t.
80,000 = 40,000(1.11)
t
Using the calculator solution method from Explain 2, t ≈ 6.64.
The population will reach 80,000 in approximately 6.6 years.
19. A lake has a small population of a rare endangered fish. The lake currently has a
population of 10 fish. The number of fish is increasing at a rate of 4% per year. When
will the population double? How long will it take the population to be 80 fish?
t
The model for population is P(t) = 10(1.04) , where t is the time in years
and P(t) is the population in year t.
Solve P(t) = 20 for t.
20 = 10(1.04)
t
Using the calculator solution method from Explain 2, t ≈ 17.67.
The population of the fish will double in 18 years.
To find when the population will be 80, you can solve P(t) = 80 for t.
© Houghton Mifflin Harcourt Publishing Company
Alternatively, note that 80 = 10 · 8 = 10 · 2 3. This corresponds to the
population doubling three times, from 10 to 20, from 20 to 40, and from
40 to 80. The population will be 80 in 54 years (3 · 18).
20. Tim has a savings account with the bank. The bank pays him 1% per year. He has
$5000 and wonders when it will reach $5200. When will his savings reach $5200?
The model is S(t) = 5000(1.01) . Solve S(t) = 5200 for t.
t
5200 = 5000(1.01)
t
Using the calculator solution method from Explain 2, t ≈ 3.94.
Graphing f(x) and g(x), we get the point of intersection (3.941648, 1.04).
Rounding up and considering interest is calculated yearly, it will take Tim
4 years.
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Lesson 1
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21. Tim is considering a different savings account that pays 1%, but this time it is
compounded monthly.
AVOID COMMON ERRORS
Some students may be unsure how to raise a
fraction to a power. Remind them that both the
numerator and the denominator must be raised to
the same power.
(When interest is compounded monthly, the bank pays interest every month instead
nt
of every year. The function representing compounded interest is S(t) = P(1 + __nr ) ,
where P is the principal, or initial deposit in the account, r is the interest rate, n is
the number of times the interest is compounded per year, t is the year, and S(t) is the
savings after t years.)
How many years will it take Tim to earn $200 at this bank? Should he switch?
(
0.01
The model is S(t) = 5000 1 + ___
12
5200 = 5000(1.00083)
12t
)
12t
12t
or S(t) = 5000(1.00083) . Solve S(t) = 5200 for t.
CURRICULUM INTEGRATION
Encourage students to research applications of
exponential functions. They should consider
applications in science and business as well as uses in
other math courses.
Using the calculator solution method from Explain 2, t ≈ 3.92.
t is approximately 4 years. Both accounts will reach $5200 in about 4 years.
Switching won’t make much difference.
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Reasoning
22. Lisa has a credit card that charges 3% interest on a monthly balance. She buys a $200
bike and plans to pay for it by making monthly payments of $100. How many months
will it take her to pay it off? Assume the first payment she makes is charged no
interest because she paid it before the first bill.
As students solve real-world problems involving time,
have them make predictions before calculating their
results. Write the predictions on the board, then
compare them to the solutions found algebraically or
graphically. Encourage students to improve their
predictions by analyzing whether their predictions
tend to be too high or too low and by considering
how they can change their estimation methods.
Her first payment is $100. At that time she owes $100 plus interest or $103. The second
month she pays $100 and the third month she pays the rest. It takes her three months to
pay it off. You do not have to solve an exponential because 3% is not a very high interest.
23. Analyze Relationships A city has 175,000 residents. The population is increasing
at the rate of 10% per year.
© Houghton Mifflin Harcourt Publishing Company
a. You want to make a prediction for how long it will take for the population to
reach 300,000. Round your answer to the nearest tenth of a year.
b. Suppose there are 350,000 residents of another city. The population of this city
is decreasing at a rate of 3% per year. Which city’s population will reach 300,000
sooner? Explain.
a. 300,000 = 175,000(1.1)
x
Using the calculator solution method from Explain 2, x ≈ 5.7.
The population will reach 300,000 in approximately 5.7 years.
b. 300,000 = 350,000(0.97)
x
Using the calculator solution method from Explain 2, x ≈ 5.1.
5.1 < 5.7
The second city’s population will reach 300,000 sooner.
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Lesson 1
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Using Graphs and Properties to Solve Equations with Exponents
614
VISUAL CUES
H.O.T. Focus on Higher Order Thinking
24. Explain the Error Jean and Marco each solved the equation 9(3) = 729. Whose
solution is incorrect? Explain your reasoning. How could the person who is incorrect
fix the work?
x
Have students create posters as visual reminders of
how to solve equations involving exponents.
Remind students to include examples as well as
step-by-step procedures.
Jean
Marco
9(3) = 729
9(3) = 729
x
x
(_91 ) ⋅ 9(3) = (_19 ) ⋅ 729
3 2 ⋅ (3) = 729
x
x
3 x = 81 = 3 4
JOURNAL
3 2 + x = 729 = 3 6
x=4
In their journals, have students explain how to use
the Equality of Bases Property to solve an equation
with a variable exponent.
x=6
Jean is completely correct and Marco could correct his work as follows:
Marco
9(3) = 729
x
3 2 ⋅ (3) = 729
x
2+x
3
= 729 = 3 6
x+2=6
x=4
He substituted for x = 6 instead of x + 2 = 6, which yields x = 4.
25. Critical Thinking Without solving, state the column containing the equation with
the greater solution for each pair of equations. Explain your reasoning.
1 (3) x = 243
1 (9) x = 243
_
_
3
3
()
()
1( )
The equation _
3 = 243 has a greater solution. Since the values of the
3
x
powers of 3 increase less quickly than the values of the powers of 9, the
© Houghton Mifflin Harcourt Publishing Company
1( )
1( )
value of x in _
3 = 243 will be greater than the value of x in _
9 = 243.
3
3
x
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Lesson 13.1
x
615
Lesson 1
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Lesson Performance Task
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Modeling
A town has a population of 78,918 residents. The town council
is offering a prize for the best prediction of how long it will take
the population to reach 100,000. The population rate is increasing
6% per year. Find the best prediction in order to win the prize. Write
an exponential equation in the form y = abx and explain what a and b
represent.
Before students write an equation for the situation in
the Lesson Performance Task, discuss how they know
that the base to be raised to a power in the
exponential equation is 1.06 and not 0.06. Have them
consider “What factor multiplied by the population
makes the number 6% greater?” Then ask what the
base would be if the population were decreasing by
6% per year. Students should recognize that it would
be 1 – 0.06 = 0.94. Discuss what a graph showing
each growth rate would look like.
Write an exponential equation.
Let y represent the population and x represent time
in years.
a represents the initial population, 78,918.
b represents the rate of increase in the population per year.
y = 78,918(1 + 0.06)
x
Substitute the target population for y: 100,000 = 78,918(1 + 0.06)
x
Write the expressions from the two sides of the equation as functions.
f(x) = 100,000
g(x) = 78,918(1 + 0.06)
x
Use the intersect feature on a graphing calculator to find the point of intersection.
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Technology
The point of intersection is (4.063245, 100,000).
The population will reach 100,000 in just over 4 years.
© Houghton Mifflin Harcourt Publishing Company · Image Credits ©Jim West/
Alamy Images
Module 13
616
As students use their graphing calculators to
graph the two functions and find their
intersection, remind them to adjust the viewing
window so that the intersection is shown clearly.
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Communication
Have students share their reasons for why the point
where the graphs of the right- and left-hand sides of
the equation intersect is the solution.
Lesson 1
EXTENSION ACTIVITY
A1_MTXESE353879_U5M13L1.indd 616
Have students research the current population of their community or state and the
rate at which it is growing or decreasing. Then have students write an exponential
equation in which y represents the population and x represents time in years.
Finally, have students choose a future population size and predict when the
population will reach that size.
13/11/14 4:46 PM
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Using Graphs and Properties to Solve Equations with Exponents
616