Download LectureSection4.3_1

4.3 Using Exponential Equations
 Given a  y , if I am looking for “y”….no big deal, just tell me what “a”
x
equals and what “x” equals.

23  8
Given a  y , if I am looking for “x” and I know what “a” and “y” it’s not
x
2  15
always so easy.
The answer is going to be somewhere between x = 3 and x = 4 (closer to 4).
We need a function related to exponents. If we are looking for the exponent and
know everything else, we can find it. That mathematical function is called a
logarithm.
x
log 2 15  2
2 x  15
The log, base 2, of 15 equals “x”.
This whole concept is similar to when you first learned about radicals
(square root, cubic roots, etc.) They were basically the inverse of
Where is
that
exponents. We could use them to “undo” an exponent.
exponent?
Logarithms are used when you are LOOKING FOR THE EXPONENT
Introduction to logarithms
https://www.khanacademy.org/math/algebra2/exponential-andlogarithmic-functions/introduction-to-logarithms/v/logarithms
There are lots of properties/rules for using exponents. We will talk
about one…The Power Rule
Let x  log a M
is equivalent to
ax  M
ax   M p
p
Now raise both side to the power of “p”
a xp  M p
Now change back to a log
Substitute log a M in for x.
Change the order of the multiplication
on the right side and we get
log a M p  xp
`
equivalent
log a M p  log a Mp
log a M p  p log a M
When you take the log of any value raised to a power it equals that power times
the log of the number.
Your book give you a short cut, but I wanted you to know why.
2x  8
log 8
x
log 2
x3
Example 4.3.1: Solving an Exponential Equation
Suppose we want to model the decrease in a river’s flow in the summer to predict when it will be 300
CFS. We observe the level to be 1860 CFS (cubic feet per second) on Monday, 1702 CFS on Tuesday, and
1557 CFS on Wednesday.
Recall from section 4.1 that exponential equations take the form y = abx; where “a” represents the initial
population, “b” represents the rate of growth, and “x” is the time.
Solution:
1702
1557
Notice that
≈ 0.915 and
≈ 0.915 so you could say the level is dropping around 8.5% per day, or
1860
1702
each day you have about 91.5% of what you had the day before.
The exponential equation F = 1860•0.915d is a good model since the level is dropping at a consistent
rate; where F = flow and d = day.
We want to know what day the flow will drop to 300 CFS.
F = 1860•0.915d
300 = 1860•0.915d
replace F with 300
d
.1613 ≈ 0.915
divide both sides by 1860
d ≈ Log 0.915 0.1613
d ≈ (Log 0.1613)/(log0.915)
introduce logarithms to solve for d
d ≈ 20.5
enter Log(0.1613)/Log(0.915) in your calculator
Final Answer: If the pattern continues, the level will hit 300 CFS about half way through day 20. This
would be on Sunday about 3 weeks away.
Note: It is important to recognize that growth or decline is always measured relative to
100%. If a population of 490 elk is increasing at 13% the model would be
y = 490(100%+13%)x or y = 490•1.13x. If the population was decreasing at 13% the model
would be y = 490(100%-13%)x or y = 490•0.87x.
We could have also entered our data into the calculator and found our model using
exponential regression. Remember that regression is better when the percent decrease is
less consistent since regression will take an average.
0
1860
1
1702
2
1557
Sample Problem 4.3.2
The Southern White Rhino, considered extinct in
the 1800’s, has enjoyed a steady increase in
population in the last hundred years.
a) Consider 1929 to be year 0 and use
regression to find an exponential
equation to model the data. Round the
numbers in your equation to 3 decimal
places.
b) It is clear in the table that the population
reached 1000 between 1948 and 1968. Use your equation to find the year, accurate to the
nearest tenth place.
c) Use your equation to predict the year the population will reach
Years
30,000, accurate to the nearest tenth place.
Yea r
Since
Population
1929
0
19
39
58
68
80
81
Solutions:
a)
Enter into the calculator the data from the table, then select the
1929
1948
1968
1987
1997
2009
2010
150
550
1800
4665
8440
11640
20170
ExpReg option to generate the regression equation: y  164.466 1.061
x
b)
Using the equation from a),
c)
Using the equation from a),
substitute 1,000 for y and solve for x.
substitute 30,000 for y and solve for x.
1000  164.466  1.061x
1000
 1.061x
164.466
6.0802  1.061x
30000  164.466  1.061x
30000
 1.061x
164.466
182.4085  1.061x
log 6.0802  log1.061x
log 6.0802  x log1.061
log182.4085  log1.061x
log182.4085  x log1.061
log 6.0802
log1.061
x  30.5
x  year 1959.5 (1929  30.5)
x
x
log182.4085
log1.061
x  87.9
x  year 2016.9 (1929  87.9)
Sample Problem 4.3.4
Temperature and moisture content in wood are related, in
that as the temperature increases the percent moisture in the
wood decreases.
a) Use regression to find an exponential equation to model
the data. Round the numbers in your equation to 3
decimal places.
b) Use your equation to accurately find the temperature, accurate to the nearest tenth place,
needed to get the moisture content to 5%.
c) Use your equation to
predict the temperature,
accurate to the nearest
tenth place that would be
necessary to get the
moisture content to 2%.
Solutions:
a)
Enter into the calculator the data from the table, then select the ExpReg option to generate the
regression equation: y  9.834  0.972
b)
x
Using the equation from a),
substitute 5 for y and solve for x.
c)
Using the equation from a),
substitute 2 for y and solve for x.
5  9.834  0.972 x
5
 0.972 x
9.834
0.5084  0.972 x
2  9.834  0.972 x
2
 0.972 x
9.834
0.2033  0.972 x
log 0.5084  log 0.972 x
log 0.2033  log 0.972 x
log 0.2033  x log 0.972
log 0.2033
x
log 0.972
x  56.1°
log 0.5084  x log 0.972 x
log 0.5084
x
log 0.972
x  23.82°