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Fitting Exponential Models to Data

Module by: First Last. E-mail the author

Summary: In this section, you will:

  • Build an exponential model from data.
  • Build a logarithmic model from data.
  • Build a logistic model from data.

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In previous sections of this chapter, we were either given a function explicitly to graph or evaluate, or we were given a set of points that were guaranteed to lie on the curve. Then we used algebra to find the equation that fit the points exactly. In this section, we use a modeling technique called regression analysis to find a curve that models data collected from real-world observations. With regression analysis, we don’t expect all the points to lie perfectly on the curve. The idea is to find a model that best fits the data. Then we use the model to make predictions about future events.

Do not be confused by the word model. In mathematics, we often use the terms function, equation, and model interchangeably, even though they each have their own formal definition. The term model is typically used to indicate that the equation or function approximates a real-world situation.

We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key features and definitions of its associated function are included for review. Take a moment to rethink each of these functions, reflect on the work we’ve done so far, and then explore the ways regression is used to model real-world phenomena.

Building an Exponential Model from Data

As we’ve learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that’s not the whole story. It’s the way data increase or decrease that helps us determine whether it is best modeled by an exponential equation. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let’s review exponential growth and decay.

Recall that exponential functions have the form y=a b x y=a b x or y= A 0 e kx . y= A 0 e kx . When performing regression analysis, we use the form most commonly used on graphing utilities, y=a b x . y=a b x . Take a moment to reflect on the characteristics we’ve already learned about the exponential function y=a b x y=a b x (assume a>0): a>0):

  • b b must be greater than zero and not equal to one.
  • The initial value of the model is y=a. y=a.
    • If b>1, b>1, the function models exponential growth. As x x increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound.
    • If 0<b<1, 0<b<1, the function models exponential decay. As x x increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the x-axis. In other words, the outputs never become equal to or less than zero.

As part of the results, your calculator will display a number known as the correlation coefficient, labeled by the variable r, r, or r 2 . r 2 . (You may have to change the calculator’s settings for these to be shown.) The values are an indication of the “goodness of fit” of the regression equation to the data. We more commonly use the value of r 2 r 2 instead of r, r, but the closer either value is to 1, the better the regression equation approximates the data.

A General Note: Exponential Regression :

Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. We use the command “ExpReg” on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form, y=a b x y=a b x

Note that:

  • b b must be non-negative.
  • when b>1, b>1, we have an exponential growth model.
  • when 0<b<1, 0<b<1, we have an exponential decay model.

How To:

Given a set of data, perform exponential regression using a graphing utility.

  1. Use the STAT then EDIT menu to enter given data.
    1. Clear any existing data from the lists.
    2. List the input values in the L1 column.
    3. List the output values in the L2 column.
  2. Graph and observe a scatter plot of the data using the STATPLOT feature.
    1. Use ZOOM [9] to adjust axes to fit the data.
    2. Verify the data follow an exponential pattern.
  3. Find the equation that models the data.
    1. Select “ExpReg” from the STAT then CALC menu.
    2. Use the values returned for a and b to record the model, y=a b x . y=a b x .
  4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data.

Example 1

Problem 1

Using Exponential Regression to Fit a Model to Data

In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person’s blood alcohol level (BAC) with the risk of being in an accident. Table 1 shows results from the study 1. The relative risk is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol.

Table 1
BAC 0 0.01 0.03 0.05 0.07 0.09
Relative Risk of Crashing 1 1.03 1.06 1.38 2.09 3.54
BAC 0.11 0.13 0.15 0.17 0.19 0.21
Relative Risk of Crashing 6.41 12.6 22.1 39.05 65.32 99.78
  1. Let x x represent the BAC level, and let y y represent the corresponding relative risk. Use exponential regression to fit a model to these data.
  2. After 6 drinks, a person weighing 160 pounds will have a BAC of about 0.16. 0.16. How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth.
Solution
  1. Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown in Figure 1:
    Figure 1
    Graph of a scattered plot.

    Use the “ExpReg” command from the STAT then CALC menu to obtain the exponential model,

    y=0.58304829 ( 2.20720213E10 ) x y=0.58304829 ( 2.20720213E10 ) x

    Converting from scientific notation, we have:

    y=0.58304829 ( 22,072,021,300 ) x y=0.58304829 ( 22,072,021,300 ) x

    Notice that r 2 0.97 r 2 0.97 which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 2:

    Figure 2
    Graph of a scattered plot with an estimation line.
  2. Use the model to estimate the risk associated with a BAC of 0.16. 0.16. Substitute 0.16 0.16 for x x in the model and solve for y. y.

    y =0.58304829 ( 22,072,021,300 ) x Use the regression model found in part (a). =0.58304829 ( 22,072,021,300 ) 0.16 Substitute 0.16 for x. 26.35 Round to the nearest hundredth. y =0.58304829 ( 22,072,021,300 ) x Use the regression model found in part (a). =0.58304829 ( 22,072,021,300 ) 0.16 Substitute 0.16 for x. 26.35 Round to the nearest hundredth.

    If a 160-pound person drives after having 6 drinks, he or she is about 26.35 times more likely to crash than if driving while sober.

Try It:

Exercise 1

Table 2 shows a recent graduate’s credit card balance each month after graduation.

Table 2
Month 1 2 3 4 5 6 7 8
Debt ($) 620.00 761.88 899.80 1039.93 1270.63 1589.04 1851.31 2154.92
  1. Use exponential regression to fit a model to these data.
  2. If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating?
Solution
  1. The exponential regression model that fits these data is y=522.88585984 ( 1.19645256 ) x . y=522.88585984 ( 1.19645256 ) x .
  2. If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year.

Q&A:

Is it reasonable to assume that an exponential regression model will represent a situation indefinitely?

No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation).

Building a Logarithmic Model from Data

Just as with exponential functions, there are many real-world applications for logarithmic functions: intensity of sound, pH levels of solutions, yields of chemical reactions, production of goods, and growth of infants. As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward. Again, it is the way they increase or decrease that helps us determine whether a logarithmic model is best.

Recall that logarithmic functions increase or decrease rapidly at first, but then steadily slow as time moves on. By reflecting on the characteristics we’ve already learned about this function, we can better analyze real world situations that reflect this type of growth or decay. When performing logarithmic regression analysis, we use the form of the logarithmic function most commonly used on graphing utilities, y=a+bln( x ). y=a+bln( x ). For this function

  • All input values, x, x, must be greater than zero.
  • The point ( 1,a ) ( 1,a ) is on the graph of the model.
  • If b>0, b>0, the model is increasing. Growth increases rapidly at first and then steadily slows over time.
  • If b<0, b<0, the model is decreasing. Decay occurs rapidly at first and then steadily slows over time.

A General Note: Logarithmic Regression :

Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time. We use the command “LnReg” on a graphing utility to fit a logarithmic function to a set of data points. This returns an equation of the form,

y=a+bln( x ) y=a+bln( x )

Note that

  • all input values, x, x, must be non-negative.
  • when b>0, b>0, the model is increasing.
  • when b<0, b<0, the model is decreasing.

How To:

Given a set of data, perform logarithmic regression using a graphing utility.

  1. Use the STAT then EDIT menu to enter given data.
    1. Clear any existing data from the lists.
    2. List the input values in the L1 column.
    3. List the output values in the L2 column.
  2. Graph and observe a scatter plot of the data using the STATPLOT feature.
    1. Use ZOOM [9] to adjust axes to fit the data.
    2. Verify the data follow a logarithmic pattern.
  3. Find the equation that models the data.
    1. Select “LnReg” from the STAT then CALC menu.
    2. Use the values returned for a and b to record the model, y=a+bln( x ). y=a+bln( x ).
  4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data.

Example 2

Problem 1

Using Logarithmic Regression to Fit a Model to Data

Due to advances in medicine and higher standards of living, life expectancy has been increasing in most developed countries since the beginning of the 20th century.

Table 3 shows the average life expectancies, in years, of Americans from 1900–20102.

Table 3
Year 1900 1910 1920 1930 1940 1950
Life Expectancy(Years) 47.3 50.0 54.1 59.7 62.9 68.2
Year 1960 1970 1980 1990 2000 2010
Life Expectancy(Years) 69.7 70.8 73.7 75.4 76.8 78.7
  1. Let x x represent time in decades starting with x=1 x=1 for the year 1900, x=2 x=2 for the year 1910, and so on. Let y y represent the corresponding life expectancy. Use logarithmic regression to fit a model to these data.
  2. Use the model to predict the average American life expectancy for the year 2030.
Solution
  1. Using the STAT then EDIT menu on a graphing utility, list the years using values 1–12 in L1 and the corresponding life expectancy in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logarithmic pattern as shown in Figure 3:
    Figure 3
    Graph of a scattered plot.

    Use the “LnReg” command from the STAT then CALC menu to obtain the logarithmic model,

    y=42.52722583+13.85752327ln(x) y=42.52722583+13.85752327ln(x)

    Next, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 4:

    Figure 4
    Graph of a scattered plot with an estimation line.
  2. To predict the life expectancy of an American in the year 2030, substitute x=14 x=14 for the in the model and solve for y: y:
    y =42.52722583+13.85752327ln(x) Use the regression model found in part (a). =42.52722583+13.85752327ln(14) Substitute 14 for x. 79.1 Round to the nearest tenth. y =42.52722583+13.85752327ln(x) Use the regression model found in part (a). =42.52722583+13.85752327ln(14) Substitute 14 for x. 79.1 Round to the nearest tenth.

    If life expectancy continues to increase at this pace, the average life expectancy of an American will be 79.1 by the year 2030.

Try It:

Exercise 2

Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. Table 4 shows the number of games sold, in thousands, from the years 2000–2010.

Table 4
Year 2000 2001 2002 2003 2004 2005
Number Sold (thousands) 142 149 154 155 159 161
Year 2006 2007 2008 2009 2010 -
Number Sold (thousands) 163 164 164 166 167 -
  1. Let x x represent time in years starting with x=1 x=1 for the year 2000. Let y y represent the number of games sold in thousands. Use logarithmic regression to fit a model to these data.
  2. If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand.
Solution
  1. The logarithmic regression model that fits these data is y=141.91242949+10.45366573ln(x) y=141.91242949+10.45366573ln(x)
  2. If sales continue at this rate, about 171,000 games will be sold in the year 2015.

Building a Logistic Model from Data

Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or limiting value. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients.

It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic regression analysis, we use the form most commonly used on graphing utilities:

y= c 1+a e bx y= c 1+a e bx

Recall that:

  • c 1+a c 1+a is the initial value of the model.
  • when b>0, b>0, the model increases rapidly at first until it reaches its point of maximum growth rate, ( ln( a ) b , c 2 ). ( ln( a ) b , c 2 ). At that point, growth steadily slows and the function becomes asymptotic to the upper bound y=c. y=c.
  • c c is the limiting value, sometimes called the carrying capacity, of the model.

A General Note: Logistic Regression:

Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command “Logistic” on a graphing utility to fit a logistic function to a set of data points. This returns an equation of the form

y= c 1+a e bx y= c 1+a e bx

Note that

  • The initial value of the model is c 1+a . c 1+a .
  • Output values for the model grow closer and closer to y=c y=c as time increases.

How To:

Given a set of data, perform logistic regression using a graphing utility.

  1. Use the STAT then EDIT menu to enter given data.
    1. Clear any existing data from the lists.
    2. List the input values in the L1 column.
    3. List the output values in the L2 column.
  2. Graph and observe a scatter plot of the data using the STATPLOT feature.
    1. Use ZOOM [9] to adjust axes to fit the data.
    2. Verify the data follow a logistic pattern.
  3. Find the equation that models the data.
    1. Select “Logistic” from the STAT then CALC menu.
    2. Use the values returned for a, a, b, b, and c c to record the model, y= c 1+a e bx . y= c 1+a e bx .
  4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data.

Example 3

Problem 1

Using Logistic Regression to Fit a Model to Data

Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. Table 5 shows the percentage of Americans with cellular service between the years 1995 and 2012 3.

Table 5
Year Americans with Cellular Service (%) Year Americans with Cellular Service (%)
1995 12.69 2004 62.852
1996 16.35 2005 68.63
1997 20.29 2006 76.64
1998 25.08 2007 82.47
1999 30.81 2008 85.68
2000 38.75 2009 89.14
2001 45.00 2010 91.86
2002 49.16 2011 95.28
2003 55.15 2012 98.17
  1. Let x x represent time in years starting with x=0 x=0 for the year 1995. Let y y represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data.
  2. Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent.
  3. Discuss the value returned for the upper limit, c. c. What does this tell you about the model? What would the limiting value be if the model were exact?
Solution
  1. Using the STAT then EDIT menu on a graphing utility, list the years using values 0–15 in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logistic pattern as shown in Figure 5:
    Figure 5
    Graph of a scattered plot.

    Use the “Logistic” command from the STAT then CALC menu to obtain the logistic model,

    y= 105.7379526 1+6.88328979 e 0.2595440013x y= 105.7379526 1+6.88328979 e 0.2595440013x

    Next, graph the model in the same window as shown in Figure 6 the scatterplot to verify it is a good fit:

    Figure 6
    Graph of a scattered plot with an estimation line.
  2. To approximate the percentage of Americans with cellular service in the year 2013, substitute x=18 x=18 for the in the model and solve for y: y:

    y = 105.7379526 1+6.88328979 e 0.2595440013x Use the regression model found in part (a). = 105.7379526 1+6.88328979 e 0.2595440013(18) Substitute 18 for x. 99.3  Round to the nearest tenth y = 105.7379526 1+6.88328979 e 0.2595440013x Use the regression model found in part (a). = 105.7379526 1+6.88328979 e 0.2595440013(18) Substitute 18 for x. 99.3  Round to the nearest tenth

    According to the model, about 98.8% of Americans had cellular service in 2013.

  3. The model gives a limiting value of about 105. This means that the maximum possible percentage of Americans with cellular service would be 105%, which is impossible. (How could over 100% of a population have cellular service?) If the model were exact, the limiting value would be c=100 c=100 and the model’s outputs would get very close to, but never actually reach 100%. After all, there will always be someone out there without cellular service!

Try It:

Exercise 3

Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.

Table 6
Year Seal Population (Thousands) Year Seal Population (Thousands)
1997 3.493 2005 19.590
1998 5.282 2006 21.955
1999 6.357 2007 22.862
2000 9.201 2008 23.869
2001 11.224 2009 24.243
2002 12.964 2010 24.344
2003 16.226 2011 24.919
2004 18.137 2012 25.108
  1. Let x x represent time in years starting with x=0 x=0 for the year 1997. Let y y represent the number of seals in thousands. Use logistic regression to fit a model to these data.
  2. Use the model to predict the seal population for the year 2020.
  3. To the nearest whole number, what is the limiting value of this model?
Solution
  1. The logistic regression model that fits these data is y= 25.65665979 1+6.113686306 e 0.3852149008x . y= 25.65665979 1+6.113686306 e 0.3852149008x .
  2. If the population continues to grow at this rate, there will be about 25,634 25,634 seals in 2020.
  3. To the nearest whole number, the carrying capacity is 25,657.

Media:

Access this online resource for additional instruction and practice with exponential function models.

Visit this website for additional practice questions from Learningpod.

Key Concepts

  • Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero.
  • We use the command “ExpReg” on a graphing utility to fit function of the form y=a b x y=a b x to a set of data points. See Example 1.
  • Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time.
  • We use the command “LnReg” on a graphing utility to fit a function of the form y=a+bln( x ) y=a+bln( x ) to a set of data points. See Example 2.
  • Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit.
  • We use the command “Logistic” on a graphing utility to fit a function of the form y= c 1+a e bx y= c 1+a e bx to a set of data points. See Example 3.

Section Exercises

Verbal

Exercise 4

What situations are best modeled by a logistic equation? Give an example, and state a case for why the example is a good fit.

Solution

Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic model best describes populations.

Exercise 5

What is a carrying capacity? What kind of model has a carrying capacity built into its formula? Why does this make sense?

Exercise 6

What is regression analysis? Describe the process of performing regression analysis on a graphing utility.

Solution

Regression analysis is the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined, select the appropriate regression analysis command from the STAT then CALC menu.

Exercise 7

What might a scatterplot of data points look like if it were best described by a logarithmic model?

Exercise 8

What does the y-intercept on the graph of a logistic equation correspond to for a population modeled by that equation?

Solution

The y-intercept on the graph of a logistic equation corresponds to the initial population for the population model.

Graphical

For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 7 through Figure 11. Answer using the letter beneath the matching graph.

Figure 7
Graph of a scattered plot.
Figure 8
Graph of a scattered plot.
Figure 9
Graph of a scattered plot.
Figure 10
Graph of a scattered plot.
Figure 11
Graph of a scattered plot.

Exercise 9

y=10.209 e 0.294x y=10.209 e 0.294x

Exercise 10

y=5.5981.912ln(x) y=5.5981.912ln(x)

Solution

C

Exercise 11

y=2.104 ( 1.479 ) x y=2.104 ( 1.479 ) x

Exercise 12

y=4.607+2.733ln(x) y=4.607+2.733ln(x)

Solution

B

Exercise 13

y= 14.005 1+2.79 e 0.812x y= 14.005 1+2.79 e 0.812x

Numeric

Exercise 14

To the nearest whole number, what is the initial value of a population modeled by the logistic equation P(t)= 175 1+6.995 e 0.68t ? P(t)= 175 1+6.995 e 0.68t ? What is the carrying capacity?

Solution

P(0)=22 P(0)=22 ; 175

Exercise 15

Rewrite the exponential model A(t)=1550 ( 1.085 ) x A(t)=1550 ( 1.085 ) x as an equivalent model with base e. e. Express the exponent to four significant digits.

Exercise 16

A logarithmic model is given by the equation h(p)=67.6825.792ln( p ). h(p)=67.6825.792ln( p ). To the nearest hundredth, for what value of p p does h(p)=62? h(p)=62?

Solution

p2.67 p2.67

Exercise 17

A logistic model is given by the equation P(t)= 90 1+5 e 0.42t . P(t)= 90 1+5 e 0.42t . To the nearest hundredth, for what value of t does P(t)=45? P(t)=45?

Exercise 18

What is the y-intercept on the graph of the logistic model given in the previous exercise?

Solution

y-intercept: ( 0,15 ) ( 0,15 )

Technology

For the following exercises, use this scenario: The population P P of a koi pond over x x months is modeled by the function P(x)= 68 1+16 e 0.28x . P(x)= 68 1+16 e 0.28x .

Exercise 19

Graph the population model to show the population over a span of 3 3 years.

Exercise 20

What was the initial population of koi?

Solution

4 4 koi

Exercise 21

How many koi will the pond have after one and a half years?

Exercise 22

How many months will it take before there are 20 20 koi in the pond?

Solution

about 6.8 6.8 months.

Exercise 23

Use the intersect feature to approximate the number of months it will take before the population of the pond reaches half its carrying capacity.

Solution
Graph of the intersection of P(t)=68/(1+16e^(-0.28t)) and y=34.

For the following exercises, use this scenario: The population P P of an endangered species habitat for wolves is modeled by the function P(x)= 558 1+54.8 e 0.462x , P(x)= 558 1+54.8 e 0.462x , where x x is given in years.

Exercise 24

Graph the population model to show the population over a span of 10 10 years.

Exercise 25

What was the initial population of wolves transported to the habitat?

Solution

10 10 wolves

Exercise 26

How many wolves will the habitat have after 3 3 years?

Exercise 27

How many years will it take before there are 100 100 wolves in the habitat?

Solution

about 5.4 years.

Exercise 28

Use the intersect feature to approximate the number of years it will take before the population of the habitat reaches half its carrying capacity.

For the following exercises, refer to Table 7.

Table 7
xf(x)
11125
21495
32310
43294
54650
66361

Exercise 29

Use a graphing calculator to create a scatter diagram of the data.

Solution
Graph of the table’s values.

Exercise 30

Use the regression feature to find an exponential function that best fits the data in the table.

Exercise 31

Write the exponential function as an exponential equation with base e. e.

Solution

f(x)=776.682 e 0.3549x f(x)=776.682 e 0.3549x

Exercise 32

Graph the exponential equation on the scatter diagram.

Exercise 33

Use the intersect feature to find the value of x x for which f(x)=4000. f(x)=4000.

Solution

When f(x)=4000, f(x)=4000, x4.6. x4.6.

Graph of the intersection of a scattered plot with an estimation line and y=4,000.

For the following exercises, refer to Table 8.

Table 8
xf(x)
1555
2383
3307
4210
5158
6122

Exercise 34

Use a graphing calculator to create a scatter diagram of the data.

Exercise 35

Use the regression feature to find an exponential function that best fits the data in the table.

Solution

f(x)=731.92 (0.738) x f(x)=731.92 (0.738) x

Exercise 36

Write the exponential function as an exponential equation with base e. e.

Exercise 37

Graph the exponential equation on the scatter diagram.

Solution
Graph of a scattered plot with an estimation line.

Exercise 38

Use the intersect feature to find the value of x x for which f(x)=250. f(x)=250.

For the following exercises, refer to Table 9.

Table 9
xf(x)
15.1
26.3
37.3
47.7
58.1
68.6

Exercise 39

Use a graphing calculator to create a scatter diagram of the data.

Solution
Graph of the table’s values.

Exercise 40

Use the LOGarithm option of the REGression feature to find a logarithmic function of the form y=a+bln( x ) y=a+bln( x ) that best fits the data in the table.

Exercise 41

Use the logarithmic function to find the value of the function when x=10. x=10.

Solution

f(10)9.5 f(10)9.5

Exercise 42

Graph the logarithmic equation on the scatter diagram.

Exercise 43

Use the intersect feature to find the value of x x for which f(x)=7. f(x)=7.

Solution

When f(x)=7, f(x)=7, x2.7. x2.7.

Graph of the intersection of a scattered plot with an estimation line and y=7.

For the following exercises, refer to Table 10.

Table 10
xf(x)
17.5
26
35.2
44.3
53.9
63.4
73.1
82.9

Exercise 44

Use a graphing calculator to create a scatter diagram of the data.

Exercise 45

Use the LOGarithm option of the REGression feature to find a logarithmic function of the form y=a+bln( x ) y=a+bln( x ) that best fits the data in the table.

Solution

f(x)=7.5442.268ln(x) f(x)=7.5442.268ln(x)

Exercise 46

Use the logarithmic function to find the value of the function when x=10. x=10.

Exercise 47

Graph the logarithmic equation on the scatter diagram.

Solution
Graph of a scattered plot with an estimation line.

Exercise 48

Use the intersect feature to find the value of x x for which f(x)=8. f(x)=8.

For the following exercises, refer to Table 11.

Table 11
xf(x)
18.7
212.3
315.4
418.5
520.7
622.5
723.3
824
924.6
1024.8

Exercise 49

Use a graphing calculator to create a scatter diagram of the data.

Solution
Graph of the table’s values.

Exercise 50

Use the LOGISTIC regression option to find a logistic growth model of the form y= c 1+a e bx y= c 1+a e bx that best fits the data in the table.

Exercise 51

Graph the logistic equation on the scatter diagram.

Solution
Graph of a scattered plot with an estimation line.

Exercise 52

To the nearest whole number, what is the predicted carrying capacity of the model?

Exercise 53

Use the intersect feature to find the value of x x for which the model reaches half its carrying capacity.

Solution

When f(x)=12.5, f(x)=12.5, x2.1. x2.1.

Graph of the intersection of a scattered plot with an estimation line and y=12.

For the following exercises, refer to Table 12.

Table 12
xxf(x)f(x)
012
228.6
452.8
570.3
799.9
8112.5
10125.8
11127.9
15135.1
17135.9

Exercise 54

Use a graphing calculator to create a scatter diagram of the data.

Exercise 55

Use the LOGISTIC regression option to find a logistic growth model of the form y= c 1+a e bx y= c 1+a e bx that best fits the data in the table.

Solution

f(x)= 136.068 1+10.324 e 0.480x f(x)= 136.068 1+10.324 e 0.480x

Exercise 56

Graph the logistic equation on the scatter diagram.

Exercise 57

To the nearest whole number, what is the predicted carrying capacity of the model?

Solution

about 136 136

Exercise 58

Use the intersect feature to find the value of x x for which the model reaches half its carrying capacity.

Extensions

Exercise 59

Recall that the general form of a logistic equation for a population is given by P(t)= c 1+a e bt , P(t)= c 1+a e bt , such that the initial population at time t=0 t=0 is P(0)= P 0 . P(0)= P 0 . Show algebraically that cP(t) P(t) = c P 0 P 0 e bt . cP(t) P(t) = c P 0 P 0 e bt .

Solution

Working with the left side of the equation, we see that it can be rewritten as a e bt : a e bt :

cP(t) P(t) = c c 1+a e bt c 1+a e bt = c( 1+a e bt )c 1+a e bt c 1+a e bt = c( 1+a e bt 1 ) 1+a e bt c 1+a e bt =1+a e bt 1=a e bt cP(t) P(t) = c c 1+a e bt c 1+a e bt = c( 1+a e bt )c 1+a e bt c 1+a e bt = c( 1+a e bt 1 ) 1+a e bt c 1+a e bt =1+a e bt 1=a e bt

Working with the right side of the equation we show that it can also be rewritten as a e bt . a e bt . But first note that when t=0, t=0, P 0 = c 1+a e b(0) = c 1+a . P 0 = c 1+a e b(0) = c 1+a . Therefore,

c P 0 P 0 e bt = c c 1+a c 1+a e bt = c( 1+a )c 1+a c 1+a e bt = c( 1+a1 ) 1+a c 1+a e bt =( 1+a1 ) e bt =a e bt c P 0 P 0 e bt = c c 1+a c 1+a e bt = c( 1+a )c 1+a c 1+a e bt = c( 1+a1 ) 1+a c 1+a e bt =( 1+a1 ) e bt =a e bt

Thus, cP(t) P(t) = c P 0 P 0 e bt . cP(t) P(t) = c P 0 P 0 e bt .

Exercise 60

Use a graphing utility to find an exponential regression formula f(x) f(x) and a logarithmic regression formula g(x) g(x) for the points ( 1.5,1.5 ) ( 1.5,1.5 ) and ( 8.5, 8.5 ). ( 8.5, 8.5 ). Round all numbers to 6 decimal places. Graph the points and both formulas along with the line y=x y=x on the same axis. Make a conjecture about the relationship of the regression formulas.

Exercise 61

Verify the conjecture made in the previous exercise. Round all numbers to six decimal places when necessary.

Solution

First rewrite the exponential with base e: f(x)=1.034341 e 0.247800x . f(x)=1.034341 e 0.247800x . Then test to verify that f(g(x))=x, f(g(x))=x, taking rounding error into consideration:

g(f(x)) =4.035510ln( 1.034341 e 0.247800x )0.136259 =4.03551( ln( 1.034341 )+ln( e 0.2478x ) )0.136259 =4.03551( ln( 1.034341 )+0.2478x )0.136259 =0.136257+0.999999x0.136259 =0.000002+0.999999x 0+x =x g(f(x)) =4.035510ln( 1.034341 e 0.247800x )0.136259 =4.03551( ln( 1.034341 )+ln( e 0.2478x ) )0.136259 =4.03551( ln( 1.034341 )+0.2478x )0.136259 =0.136257+0.999999x0.136259 =0.000002+0.999999x 0+x =x

Exercise 62

Find the inverse function f 1 ( x ) f 1 ( x ) for the logistic function f(x)= c 1+a e bx . f(x)= c 1+a e bx . Show all steps.

Exercise 63

Use the result from the previous exercise to graph the logistic model P(t)= 20 1+4 e 0.5t P(t)= 20 1+4 e 0.5t along with its inverse on the same axis. What are the intercepts and asymptotes of each function?

Solution
Graph of P(t)=20/(1+40.5e^(-0.5t)) and P(t)=(ln(4)-ln((20/t)-1)/0.5.

The graph of P(t) P(t)has a y-intercept at (0, 4) and horizontal asymptotes at y = 0 and y = 20. The graph of P 1 (t) P 1 (t)has an x- intercept at (4, 0) and vertical asymptotes at x = 0 and x = 20.

Chapter Review Exercises

Exponential Functions

Exercise 64

Determine whether the function y=156 ( 0.825 ) t y=156 ( 0.825 ) t represents exponential growth, exponential decay, or neither. Explain

Solution

exponential decay; The growth factor, 0.825, 0.825, is between 0 0 and 1. 1.

Exercise 65

The population of a herd of deer is represented by the function A(t)=205 (1.13) t , A(t)=205 (1.13) t , where t t is given in years. To the nearest whole number, what will the herd population be after 6 6 years?

Exercise 66

Find an exponential equation that passes through the points (2, 2.25) (2, 2.25) and (5,60.75). (5,60.75).

Solution

y=0.25 ( 3 ) x y=0.25 ( 3 ) x

Exercise 67

Determine whether Table 13 could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points.

Table 13
x 1 2 3 4
f(x) 3 0.9 0.27 0.081

Exercise 68

A retirement account is opened with an initial deposit of $8,500 and earns 8.12% 8.12% interest compounded monthly. What will the account be worth in 20 20 years?

Solution

$42,888.18 $42,888.18

Exercise 69

Hsu-Mei wants to save $5,000 for a down payment on a car. To the nearest dollar, how much will she need to invest in an account now with 7.5% 7.5% APR, compounded daily, in order to reach her goal in 3 3 years?

Exercise 70

Does the equation y=2.294 e 0.654t y=2.294 e 0.654t represent continuous growth, continuous decay, or neither? Explain.

Solution

continuous decay; the growth rate is negative.

Exercise 71

Suppose an investment account is opened with an initial deposit of $10,500 $10,500 earning 6.25% 6.25% interest, compounded continuously. How much will the account be worth after 25 25 years?

Graphs of Exponential Functions

Exercise 72

Graph the function f(x)=3.5 ( 2 ) x . f(x)=3.5 ( 2 ) x . State the domain and range and give the y-intercept.

Solution

domain: all real numbers; range: all real numbers strictly greater than zero; y-intercept: (0, 3.5);

Graph of f(x)=3.5(2^x)

Exercise 73

Graph the function f(x)=4 ( 1 8 ) x f(x)=4 ( 1 8 ) x and its reflection about the y-axis on the same axes, and give the y-intercept.

Exercise 74

The graph of f(x)= 6.5 x f(x)= 6.5 x is reflected about the y-axis and stretched vertically by a factor of 7. 7. What is the equation of the new function, g(x)? g(x)? State its y-intercept, domain, and range.

Solution

g(x)=7 ( 6.5 ) x ; g(x)=7 ( 6.5 ) x ; y-intercept: (0, 7); (0, 7); Domain: all real numbers; Range: all real numbers greater than 0. 0.

Exercise 75

The graph below shows transformations of the graph of f(x)= 2 x . f(x)= 2 x . What is the equation for the transformation?

Graph of f(x)=2^x

Logarithmic Functions

Exercise 76

Rewrite log 17 ( 4913 )=x log 17 ( 4913 )=x as an equivalent exponential equation.

Solution

17 x =4913 17 x =4913

Exercise 77

Rewrite ln( s )=t ln( s )=tas an equivalent exponential equation.

Exercise 78

Rewrite a 2 5 =b a 2 5 =b as an equivalent logarithmic equation.

Solution

log a b= 2 5 log a b= 2 5

Exercise 79

Rewrite e 3.5 =h e 3.5 =h as an equivalent logarithmic equation.

Exercise 80

Solve for x log 64 (x)= 1 3 x log 64 (x)= 1 3 to exponential form.

Solution

x= 64 1 3 =4 x= 64 1 3 =4

Exercise 81

Evaluate log 5 ( 1 125 ) log 5 ( 1 125 )without using a calculator.

Exercise 82

Evaluate log( 0.000001 ) log( 0.000001 )without using a calculator.

Solution

log( 0.000001 )=6 log( 0.000001 )=6

Exercise 83

Evaluate log(4.005) log(4.005)using a calculator. Round to the nearest thousandth.

Exercise 84

Evaluate ln( e 0.8648 ) ln( e 0.8648 )without using a calculator.

Solution

ln( e 0.8648 )=0.8648 ln( e 0.8648 )=0.8648

Exercise 85

Evaluate ln( 18 3 ) ln( 18 3 )using a calculator. Round to the nearest thousandth.

Graphs of Logarithmic Functions

Exercise 86

Graph the function g(x)=log( 7x+21 )4. g(x)=log( 7x+21 )4.

Solution


Graph of g(x)=log(7x+21)-4.

Exercise 87

Graph the function h(x)=2ln( 93x )+1. h(x)=2ln( 93x )+1.

Exercise 88

State the domain, vertical asymptote, and end behavior of the function g(x)=ln( 4x+20 )17. g(x)=ln( 4x+20 )17.

Solution

Domain: x>5; x>5; Vertical asymptote: x=5; x=5; End behavior: as x 5 + ,f(x) x 5 + ,f(x) and as x,f(x). x,f(x).

Logarithmic Properties

Exercise 89

Rewrite ln( 7r11st ) ln( 7r11st ) in expanded form.

Exercise 90

Rewrite log 8 ( x )+ log 8 ( 5 )+ log 8 ( y )+ log 8 ( 13 ) log 8 ( x )+ log 8 ( 5 )+ log 8 ( y )+ log 8 ( 13 )in compact form.

Solution

log 8 ( 65xy ) log 8 ( 65xy )

Exercise 91

Rewrite log m ( 67 83 ) log m ( 67 83 ) in expanded form.

Exercise 92

Rewrite ln( z )ln( x )ln( y ) ln( z )ln( x )ln( y ) in compact form.

Solution

ln( z xy ) ln( z xy )

Exercise 93

Rewrite ln( 1 x 5 ) ln( 1 x 5 ) as a product.

Exercise 94

Rewrite log y ( 1 12 ) log y ( 1 12 )as a single logarithm.

Solution

log y ( 12 ) log y ( 12 )

Exercise 95

Use properties of logarithms to expand log( r 2 s 11 t 14 ). log( r 2 s 11 t 14 ).

Exercise 96

Use properties of logarithms to expand ln( 2b b+1 b1 ). ln( 2b b+1 b1 ).

Solution

ln( 2 )+ln( b )+ ln( b+1 )ln( b1 ) 2 ln( 2 )+ln( b )+ ln( b+1 )ln( b1 ) 2

Exercise 97

Condense the expression 5ln( b )+ln( c )+ ln( 4a ) 2 5ln( b )+ln( c )+ ln( 4a ) 2 to a single logarithm.

Exercise 98

Condense the expression 3 log 7 v+6 log 7 w log 7 u 3 3 log 7 v+6 log 7 w log 7 u 3 to a single logarithm.

Solution

log 7 ( v 3 w 6 u 3 ) log 7 ( v 3 w 6 u 3 )

Exercise 99

Rewrite log 3 ( 12.75 ) log 3 ( 12.75 ) to base e. e.

Exercise 100

Rewrite 5 12x17 =125 5 12x17 =125 as a logarithm. Then apply the change of base formula to solve for x x using the common log. Round to the nearest thousandth.

Solution

x= log( 125 ) log( 5 ) +17 12 = 5 3 x= log( 125 ) log( 5 ) +17 12 = 5 3

Exponential and Logarithmic Equations

Exercise 101

Solve 216 3x 216 x = 36 3x+2 216 3x 216 x = 36 3x+2 by rewriting each side with a common base.

Exercise 102

Solve 125 ( 1 625 ) x3 = 5 3 125 ( 1 625 ) x3 = 5 3 by rewriting each side with a common base.

Solution

x=3 x=3

Exercise 103

Use logarithms to find the exact solution for 7 17 9x 7=49. 7 17 9x 7=49. If there is no solution, write no solution.

Exercise 104

Use logarithms to find the exact solution for 3 e 6n2 +1=60. 3 e 6n2 +1=60. If there is no solution, write no solution.

Solution

no solution

Exercise 105

Find the exact solution for 5 e 3x 4=6 5 e 3x 4=6 . If there is no solution, write no solution.

Exercise 106

Find the exact solution for 2 e 5x2 9=56. 2 e 5x2 9=56. If there is no solution, write no solution.

Solution

no solution

Exercise 107

Find the exact solution for 5 2x3 = 7 x+1 . 5 2x3 = 7 x+1 . If there is no solution, write no solution.

Exercise 108

Find the exact solution for e 2x e x 110=0. e 2x e x 110=0. If there is no solution, write no solution.

Solution

x=ln( 11 ) x=ln( 11 )

Exercise 109

Use the definition of a logarithm to solve. 5 log 7 ( 10n )=5. 5 log 7 ( 10n )=5.

Exercise 110

47. Use the definition of a logarithm to find the exact solution for 9+6ln( a+3 )=33. 9+6ln( a+3 )=33.

Solution

a= e 4 3 a= e 4 3

Exercise 111

Use the one-to-one property of logarithms to find an exact solution for log 8 ( 7 )+ log 8 ( 4x )= log 8 ( 5 ). log 8 ( 7 )+ log 8 ( 4x )= log 8 ( 5 ). If there is no solution, write no solution.

Exercise 112

Use the one-to-one property of logarithms to find an exact solution for ln( 5 )+ln( 5 x 2 5 )=ln( 56 ). ln( 5 )+ln( 5 x 2 5 )=ln( 56 ). If there is no solution, write no solution.

Solution

x=± 9 5 x=± 9 5

Exercise 113

The formula for measuring sound intensity in decibels D D is defined by the equation D=10log( I I 0 ), D=10log( I I 0 ), where I I is the intensity of the sound in watts per square meter and I 0 = 10 12 I 0 = 10 12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a large orchestra with a sound intensity of 6.3 10 3 6.3 10 3 watts per square meter?

Exercise 114

The population of a city is modeled by the equation P(t)=256,114 e 0.25t P(t)=256,114 e 0.25t where t t is measured in years. If the city continues to grow at this rate, how many years will it take for the population to reach one million?

Solution

about 5.45 5.45 years

Exercise 115

Find the inverse function f 1 f 1 for the exponential function f( x )=2 e x+1 5. f( x )=2 e x+1 5.

Exercise 116

Find the inverse function f 1 f 1 for the logarithmic function f( x )=0.25 log 2 ( x 3 +1 ). f( x )=0.25 log 2 ( x 3 +1 ).

Solution

f 1 ( x )= 2 4x 1 3 f 1 ( x )= 2 4x 1 3

Exponential and Logarithmic Models

For the following exercises, use this scenario: A doctor prescribes 300 300 milligrams of a therapeutic drug that decays by about 17% 17% each hour.

Exercise 117

To the nearest minute, what is the half-life of the drug?

Exercise 118

Write an exponential model representing the amount of the drug remaining in the patient’s system after t t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 24 24 hours. Round to the nearest hundredth of a gram.

Solution

f(t)=300 ( 0.83 ) t ;f(24)3.43g f(t)=300 ( 0.83 ) t ;f(24)3.43g

For the following exercises, use this scenario: A soup with an internal temperature of 350° 350° Fahrenheit was taken off the stove to cool in a 71°F 71°F room. After fifteen minutes, the internal temperature of the soup was 175°F. 175°F.

Exercise 119

Use Newton’s Law of Cooling to write a formula that models this situation.

Exercise 120

How many minutes will it take the soup to cool to 85°F? 85°F?

Solution

about 45 45 minutes

For the following exercises, use this scenario: The equation N( t )= 1200 1+199 e 0.625t N( t )= 1200 1+199 e 0.625t models the number of people in a school who have heard a rumor after t t days.

Exercise 121

How many people started the rumor?

Exercise 122

To the nearest tenth, how many days will it be before the rumor spreads to half the carrying capacity?

Solution

about 8.5 8.5 days

Exercise 123

What is the carrying capacity?

For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic.

Exercise 124

xf(x)
13.05
24.42
36.4
49.28
513.46
619.52
728.3
841.04
959.5
1086.28
Solution

exponential

Graph of the table’s values.

Exercise 125

xf(x)
0.518.05
117
315.33
514.55
714.04
1013.5
1213.22
1313.1
1512.88
1712.69
2012.45

Exercise 126

Find a formula for an exponential equation that goes through the points ( 2,100 ) ( 2,100 ) and ( 0,4 ). ( 0,4 ). Then express the formula as an equivalent equation with base e.

Solution

y=4 ( 0.2 ) x ; y=4 ( 0.2 ) x ; y=4 e -1.609438x y=4 e -1.609438x

Fitting Exponential Models to Data

Exercise 127

What is the carrying capacity for a population modeled by the logistic equation P(t)= 250,000 1+499 e 0.45t ? P(t)= 250,000 1+499 e 0.45t ?What is the initial population for the model?

Exercise 128

The population of a culture of bacteria is modeled by the logistic equation P(t)= 14,250 1+29 e 0.62t , P(t)= 14,250 1+29 e 0.62t , where t t is in days. To the nearest tenth, how many days will it take the culture to reach 75% 75% of its carrying capacity?

Solution

about 7.2 7.2 days

For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.

Exercise 129

xf(x)
1409.4
2260.7
3170.4
4110.6
574
644.7
732.4
819.5
912.7
108.1

Exercise 130

xf(x)
0.1536.21
0.2528.88
0.524.39
0.7518.28
116.5
1.512.99
29.91
2.258.57
2.757.23
35.99
3.54.81
Solution

logarithmic; y=16.687189.71860ln(x) y=16.687189.71860ln(x)

Graph of the table’s values.

Exercise 131

xf(x)
09
222.6
444.2
562.1
796.9
8113.4
10133.4
11137.6
15148.4
17149.3

Practice Test

Exercise 132

The population of a pod of bottlenose dolphins is modeled by the function A(t)=8 (1.17) t , A(t)=8 (1.17) t , where t t is given in years. To the nearest whole number, what will the pod population be after 3 3 years?

Solution

About 13 13 dolphins.

Exercise 133

Find an exponential equation that passes through the points (0, 4) (0, 4) and (2, 9). (2, 9).

Exercise 134

Drew wants to save $2,500 to go to the next World Cup. To the nearest dollar, how much will he need to invest in an account now with 6.25% 6.25% APR, compounding daily, in order to reach his goal in 4 4 years?

Solution

$1,947 $1,947

Exercise 135

An investment account was opened with an initial deposit of $9,600 and earns 7.4% 7.4% interest, compounded continuously. How much will the account be worth after 15 15 years?

Exercise 136

Graph the function f(x)=5 ( 0.5 ) x f(x)=5 ( 0.5 ) x and its reflection across the y-axis on the same axes, and give the y-intercept.

Solution

y-intercept: (0, 5) (0, 5)

Graph of f(-x)=5(0.5)^-x in blue and f(x)=5(0.5)^x in orange.

Exercise 137

The graph shows transformations of the graph of f(x)= ( 1 2 ) x . f(x)= ( 1 2 ) x . What is the equation for the transformation?

Graph of f(x)= (1/2)^x.

Exercise 138

Rewrite log 8.5 ( 614.125 )=a log 8.5 ( 614.125 )=a as an equivalent exponential equation.

Solution

8.5 a =614.125 8.5 a =614.125

Exercise 139

Rewrite e 1 2 =m e 1 2 =m as an equivalent logarithmic equation.

Exercise 140

Solve for x x by converting the logarithmic equation lo g 1 7 (x)=2 lo g 1 7 (x)=2 to exponential form.

Solution

x= ( 1 7 ) 2 = 1 49 x= ( 1 7 ) 2 = 1 49

Exercise 141

Evaluate log(10,000,000) log(10,000,000)without using a calculator.

Exercise 142

Evaluate ln( 0.716 ) ln( 0.716 )using a calculator. Round to the nearest thousandth.

Solution

ln( 0.716 )0.334 ln( 0.716 )0.334

Exercise 143

Graph the function g(x)=log( 126x )+3. g(x)=log( 126x )+3.

Exercise 144

State the domain, vertical asymptote, and end behavior of the function f(x)= log 5 ( 3913x )+7. f(x)= log 5 ( 3913x )+7.

Solution

Domain: x<3; x<3; Vertical asymptote: x=3; x=3; End behavior: x 3 ,f(x) x 3 ,f(x) and x,f(x) x,f(x)

Exercise 145

Rewrite log( 17a2b ) log( 17a2b )as a sum.

Exercise 146

Rewrite log t ( 96 ) log t ( 8 ) log t ( 96 ) log t ( 8 )in compact form.

Solution

log t ( 12 ) log t ( 12 )

Exercise 147

Rewrite log 8 ( a 1 b ) log 8 ( a 1 b ) as a product.

Exercise 148

Use properties of logarithm to expand ln( y 3 z 2 x4 3 ). ln( y 3 z 2 x4 3 ).

Solution

3ln( y )+2ln( z )+ ln( x4 ) 3 3ln( y )+2ln( z )+ ln( x4 ) 3

Exercise 149

Condense the expression 4ln( c )+ln( d )+ ln( a ) 3 + ln( b+3 ) 3 4ln( c )+ln( d )+ ln( a ) 3 + ln( b+3 ) 3 to a single logarithm.

Exercise 150

Rewrite 16 3x5 =1000 16 3x5 =1000 as a logarithm. Then apply the change of base formula to solve for x x using the natural log. Round to the nearest thousandth.

Solution

x= ln( 1000 ) ln( 16 ) +5 3 2.497 x= ln( 1000 ) ln( 16 ) +5 3 2.497

Exercise 151

Solve ( 1 81 ) x 1 243 = ( 1 9 ) 3x1 ( 1 81 ) x 1 243 = ( 1 9 ) 3x1 by rewriting each side with a common base.

Exercise 152

Use logarithms to find the exact solution for 9 e 10a8 5=41 9 e 10a8 5=41 . If there is no solution, write no solution.

Solution

a= ln( 4 )+8 10 a= ln( 4 )+8 10

Exercise 153

Find the exact solution for 10 e 4x+2 +5=56. 10 e 4x+2 +5=56. If there is no solution, write no solution.

Exercise 154

Find the exact solution for 5 e 4x1 4=64. 5 e 4x1 4=64. If there is no solution, write no solution.

Solution

no solution

Exercise 155

Find the exact solution for 2 x3 = 6 2x1 . 2 x3 = 6 2x1 . If there is no solution, write no solution.

Exercise 156

Find the exact solution for e 2x e x 72=0. e 2x e x 72=0. If there is no solution, write no solution.

Solution

x=ln( 9 ) x=ln( 9 )

Exercise 157

Use the definition of a logarithm to find the exact solution for 4log( 2n )7=11 4log( 2n )7=11

Exercise 158

Use the one-to-one property of logarithms to find an exact solution for log( 4 x 2 10 )+log( 3 )=log( 51 ) log( 4 x 2 10 )+log( 3 )=log( 51 ) If there is no solution, write no solution.

Solution

x=± 3 3 2 x=± 3 3 2

Exercise 159

The formula for measuring sound intensity in decibels D D is defined by the equation D=10log( I I 0 ), D=10log( I I 0 ), where I I is the intensity of the sound in watts per square meter and I 0 = 10 12 I 0 = 10 12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a rock concert with a sound intensity of 4.7 10 1 4.7 10 1 watts per square meter?

Exercise 160

A radiation safety officer is working with 112 112 grams of a radioactive substance. After 17 17 days, the sample has decayed to 80 80 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest day, what is the half-life of this substance?

Solution

f(t)=112 e .019792t ; f(t)=112 e .019792t ; half-life: about 35 35 days

Exercise 161

Write the formula found in the previous exercise as an equivalent equation with base e. e. Express the exponent to five significant digits.

Exercise 162

A bottle of soda with a temperature of 71° 71° Fahrenheit was taken off a shelf and placed in a refrigerator with an internal temperature of 35° F. 35° F. After ten minutes, the internal temperature of the soda was 63° F. 63° F. Use Newton’s Law of Cooling to write a formula that models this situation. To the nearest degree, what will the temperature of the soda be after one hour?

Solution

T(t)=36 e 0.025131t +35;T( 60 ) 43 o F T(t)=36 e 0.025131t +35;T( 60 ) 43 o F

Exercise 163

The population of a wildlife habitat is modeled by the equation P( t )= 360 1+6.2 e 0.35t , P( t )= 360 1+6.2 e 0.35t , where t t is given in years. How many animals were originally transported to the habitat? How many years will it take before the habitat reaches half its capacity?

Exercise 164

Enter the data from Table 19 into a graphing calculator and graph the resulting scatter plot. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic.

Table 19
xf(x)
13
28.55
311.79
414.09
515.88
617.33
718.57
819.64
920.58
1021.42

Solution

logarithmic

Graph of the table’s values.

Exercise 165

The population of a lake of fish is modeled by the logistic equation P(t)= 16,120 1+25 e 0.75t , P(t)= 16,120 1+25 e 0.75t , where t t is time in years. To the nearest hundredth, how many years will it take the lake to reach 80% 80% of its carrying capacity?

For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.

Exercise 166

xf(x)
120
221.6
329.2
436.4
546.6
655.7
772.6
887.1
9107.2
10138.1

Solution

exponential; y=15.10062 ( 1.24621 ) x y=15.10062 ( 1.24621 ) x

Graph of the table’s values.

Exercise 167

xf(x)
313.98
417.84
520.01
622.7
724.1
826.15
927.37
1028.38
1129.97
1231.07
1331.43

Exercise 168

xf(x)
02.2
0.52.9
13.9
1.54.8
26.4
39.3
412.3
515
616.2
717.3
817.9

Solution

logistic; y= 18.41659 1+7.54644 e 0.68375x y= 18.41659 1+7.54644 e 0.68375x

Graph of the table’s values.

Footnotes

  1. Source: Indiana University Center for Studies of Law in Action, 2007
  2. Source: Center for Disease Control and Prevention, 2013
  3. Source: The World Bank, 2013

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