Discussion Questions – Logarithmic ModelsIn Exercises 1–3 determine the doubling time for the given information, presuming the trendscontinued.1. Health care costs for U.S. employers rose 7.3% in 2018.2. Denise’s cake sales have risen 15% since last year.3. In 1991 Brazil had a population of 147,053,940 and in 1996 the population reached157,079573.Discuss the answers to Exercises 4 and 5 below:4. From 1975 to 2009 the total of atmospheric emissions of CO2 doubled. In 2010 there wereapproximately 7 billion metric tons emitted. If this trend continues, in which year will weexceed 20 billion tons of CO2 emissions?5. In March 1964 the largest earthquake ever recorded in the U.S.A. happened in PrinceWilliam Sound, Alaska. It measured 9.2 on the Richter scale. How does that compare tothe Mexico City, Mexico earthquake of February 2018, which measured 7.2 on the Richterscale? Chapter 6
Chapter 6 Part 3 – Overview
6.1
Understanding the Concept of a Function
6.2
Developing and Interpreting Linear Models
6.3
Developing and Interpreting Exponential Models
6.4
Developing and Interpreting Logarithmic Models
6.4 Developing and Interpreting Logarithmic
Models
Learning Objectives
In this section you will:
1. Use logarithmic functions to calculate the doubling time coefficient.
2. Use logarithmic functions to estimate elapsed time and predict the time of
future events.
3. Learn to manipulate exponential and logarithmic equations to answer real
world questions.
The logarithmic function alone has some uses in real life applications, but it also
allows us to use the exponential function for solving more problems. By definition the
logarithmic function is the inverse function of the exponential function. The simplest way to
think about the previous statement is to consider a graph of the exponential function in the
Cartesian Plane that is reflected about the line y = x. See figure 6.9. The domains and ranges
of the respective functions switch.
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Gallo, Harvey, & Clark, © 2011
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y
y=ax
y=x
x = ay y = logax
x
Figure 6.9 Inverse Functions
Before we begin our formal examination of logarithmic functions, we will review logs
and their interactions with exponential functions. Refer to Figure 6.10 below. All values are
real numbers where: 0 a 1 or a 1 , 0 b 1 or b 1 , c 0 , m 0 , and n 0 .
Algebraic Definition
Example
c a b log a c
8 2 3 log 2 8
log e m ln m
log e 5 ln 5
Natural log.
log 10 m log m
log 10 9 log 9
Common log.
log a mn log a m log a n
log 3 14 log 3 2 log 3 7
The input product of a log can
be written as the sum of logs.
The input ratio of a log can be
written as the difference of
logs.
The exponent is the magnitude
of the log.
b
log a
m
log a m log a n
n
3
log 7
5
log 7 5 log 7 3
3
log a m n n log a m
ln 2 5 5 ln 2
log a a 1
log 3 3 1
Adapted with permission
Gallo, Harvey, & Clark, © 2011
Explanation
Definition of a log.
Think a a .
?
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Algebraic Definition
Example
log a a m
log 10 4
log a 1 0
log 12 1 0
If log a m log a n , then m n
If log 6 30 log 6 x , then
m
4
x 30
log a m
log b m
log b a
log 2 3
log e 3
log e 2
Explanation
Combine the 2 previous
properties to see how this must
be true.
Think a 1 .
The “duh” property, or more
commonly known as the
equality property.
Change of base: most
calculators can only calculate
the natural and common logs.
?
Figure 6.10 Primer on Logs
EXAMPLE 1
Find the missing value for each problem (approximate to 4 decimal places using
a calculator where appropriate):
a. ln e 7 x
b. log 2 32 x
c. log 1000 x
d. ln 2 x
e. log 5 7 x
Solution
a. Recalling that ln a log e a , then by inspection and our knowledge of
log properties reviewed in Figure 6.10, we see that x = 7.
b. We can see that 32 can be written as 25, so we may write log 2 2 5 x .
Thus, x = 5.
c. First, recall that log a log 10 a , so log 1000 log 10 1000 . And, 1000 can
be written as 103, thus log 10 1000 log 10 10 3 . Therefore, x = 3.
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d. Because 2 is not an exponential multiple of e, we will use a calculator
to estimate the value. Before we do so, however, we need to make a
mental estimation to ensure we get the correct value and we didn’t
input something incorrectly. Recall that e 2.7 . Thus, we think
2.7 ? 2 . Clearly our answer must be less than one, so ln 2 0.6931
seems reasonable. Thus, x ≈ 0.6931.
e. In this problem, 7 is not an exponential multiple of 5 and most
calculators won’t calculate bases other than e or 10, so we must use
the change of base formula: log a b
log c b
. (Note: We can change to
log c a
either the natural log or the common log, for the ratios are the same,
but we will use the natural log for this conversion.). Our estimation
this time is 5? 7 , so we expect an answer between 1 and 2.
log 5 7
ln 7 1.9459
1.2091 . Thus, x ≈ 1.2091.
ln 5 1.6094
As mentioned in Section 6.3, most people think of exponential growth in terms of the
doubling rate. Now that we have a cursory understanding of logarithmic properties, we can
investigate and understand this phenomenon further. Consider the exponential function
f (t ) a (b) t . For our example, we will consider a growth rate of 5% per year with a
population of 1 (i.e., a = 1) at time equals zero (i.e., t = 0) and we want to know when the
population will double (i.e., f(t) = 2). Hence, 2 1(1.05) t . In order to solve an exponential
equation, we log both sides of the equation:
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Gallo, Harvey, & Clark, © 2011
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ln 2 ln(1.05) t
Equality Property
ln 2 t ln(1.05)
Exponential Property
t
ln 2
ln(1.05)
t
ln 2
14.2067
ln(1.05)
Dividing both sides of the equation by ln(1.05)
Thus, we conclude that a population growth of 5% per year would double approximately
every 14 years. A simple way to estimate the doubling time is 70 divided by the percent of
growth. For example, the doubling time of 7% growth is 10 years: 70 ÷ 7 = 10. The doubling
time of 10% growth per year is 7 years: 70 ÷ 10 = 7. The doubling time of 20% per year
growth is 3.5 years: 70 ÷ 20 = 3.5. So, a quick approximation of the doubling time for an
annual percentage of growth is
70
.
percent of growth
EXAMPLE 2
a. The U.S.A. inflation rate for 2008 was 3.8%. Had this rate remained
constant, in what year would a $2 Big Mac sandwich have cost $4?
b. Between the last two censuses in the U.S.A., the growth rate was
13% per decade. What is the doubling time in years?
Solution
a. 70 ÷ 3.8 ≈ 18.4216. So in the 19th year after 2008, we would expect
the price to double: 2027.
b. The model is 2 1(1.13)
t
for 10 years, so for every year it is:
2 1(1.13) 0.1t .
Adapted with permission
Gallo, Harvey, & Clark, © 2011
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ln 2 ln(1.13) 0.1t
ln 2 0.1t ln(1.13)
t
ln 2
0.1 ln 1.13
t 56.7131
Thus, the doubling time is approximately every 57 years.
In much of logarithmic modeling, the purpose is to rewrite an exponential function so
time is a function of quantity. For example, in our aphid model in the Week 13 Overview, the
function of aphids as predicted by time is f (t ) 2(2) .2t , f(t) is the number of aphids in t days.
If we wanted to know the T(n) years, where n is the number of aphids, we could write a
logarithmic function by using our properties of logs and properties of exponents to solve for
time:
f (t ) 2(2) .2t
Given
n 2(2) .2t
Let f(t) = n
n
(2) .2t
2
Divide both sides of equation by 2
Taking the natural log of both sides
From Figure 6.6: a mn a m
From Figure 6.10: ln a m m ln a
ln
n
ln 2 .2t
2
ln
t
n
ln 2.2
2
ln
n
t ln 2.2
2
Adapted with permission
Gallo, Harvey, & Clark, © 2011
n
Page |6
n
2
t
ln 2 .2
ln
Dividing both sides of the equation by ln 2 .2
Thus we now have a function of n number of aphids predicting time in days:
n
2
T ( n)
ln 2 .2
ln
EXAMPLE 3
The human body contains serum proteins that have a half-life of approximately
t
1
10 days. Using the formula f (t ) 128 , write a logarithmic function for
2
predicting time in T days based on n number of serum proteins (i.e., time is a
function of the number of serum proteins) and determine when there will be 4
serum proteins.
Solution
The first part of the assignment requires solving for t:
1
f (t ) 128
2
t
1
f (t ) 128
2
.1t
1
n 128
2
n
1
128 2
ln
Given for 10 days
Adjusted for per day
.1t
Let f(t) = n
.1t
n
1
ln
128
2
Dividing both sides of equation by 128
.1t
Adapted with permission
Gallo, Harvey, & Clark, © 2011
Take natural log of both sides of equation
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t
ln
1 .1
n
ln
2
128
ln
1 .1
n
t ln
128 2
t
n
ln
128
1
ln
2
From Figure 6.6: a mn a m
n
From Figure 6.10: ln a m m ln a
1
Dividing both sides of equation by ln
.1
2
.1
Thus, for the first part of the prompt: T (n)
n
ln
128
1
ln
2
.1
.
To answer the second part of the assignment, we need to find T(4):
T ( 4)
4
ln
128
1
ln
2
.1
= 50. days.
There are other times that the formulas are already in logarithmic form and all we need
to do is input the data and interpret the results. For example, the simplistic formula for
1 K f
calculating dates using the Potassium-Argon method is t 1.248 10 9 ln ln
,
2 100
where Arf is the percent of Argon isotope 40 found in the sample. Thus, if we knew that the
percent of Arf was 67, we could calculate the time at:
1 66
t 1.248 10 9 ln ln
2 100
t 3.6 10 8
In other words, the fossil would be about 360 million years old.
Adapted with permission
Gallo, Harvey, & Clark, © 2011
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EXAMPLE 4
a. The function for how old something is compared to the percent of
1 p
, where T(p) is
2 100
carbon-14 measured is T ( p) 5730 ln ln
time predicted by p percent of carbon-14 remaining in a fossil, 5730
1
2
is the number of years for carbon-14 to decay 50%, and ln ,
represents the half-life. Simplify the equation by calculating the
constant and change the parameters so only two calculations need to
be made.
b. How old is a fossil containing 30% carbon-14?
Solution
a. For this we need to calculate ln
1
and multiply it by 5730, which is
2
approximately 3971.7333 . Further, if we set the direction to T(d),
where d is the decimal value of the percent, we get the formula:
T (d ) 3971 .7333 ln d .
b. We need to mentally convert 30% to a decimal, which is .3, so we
need to find T(.3):
T (3) 3971 .7333 ln .3
T (3) 4781 .8589
Thus, the age of the fossil is approximately 4782 years.
Adapted with permission
Gallo, Harvey, & Clark, © 2011
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Key Takeaways
•
The logarithmic function is a function of magnitude.
•
The logarithmic function is the inverse of the exponential function.
•
The log of a number is simply a constant.
•
To estimate the doubling time of percentage of growth, divide 70 by
the percent.
Key Terms
Exponential Function
An equation of the model f ( x) a x where 0 a 1 or a 1 . For every unit of increase of
the domain, the range changes via a common ratio.
Inverse Function
The domain and range are reversed. For example, y a x and x a y are inverse functions.
Logarithmic Function
An equation of the model f ( x) log a x , where the relationship is c log a b a c b and
0 a 1 or a 1 . For every common ratio change of the domain, the range changes
one unit.
Adapted with permission
Gallo, Harvey, & Clark, © 2011
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