MX Algebra Fitting Exponential Models to Data

Section 6.8Fitting Exponential Models to Data

Corequisite Skills review (optional warm-up)

Learning Objectives

  • Draw and interpret scatter diagrams (linear, exponential, logarithmic). (CA 4.3.1)
  • Fit a regression equation to a set of data and use the linear (or exponential) model to make predictions. (CA 4.3.4)

Objective 1: Draw and interpret scatter diagrams (linear, exponential, logarithmic). (CA 4.3.1)

Vocabulary and Concept Check

Draw and interpret scatter diagrams (linear, exponential, logarithmic).

Fill in the blanks and match the description with the graphs a, b, or c

A ________ function has equation \(f\left(x\right)=mx+b\) and has a basic shape ________.
A ________ function has equation \(f\left(x\right)={a}^{x}, a>0, a\ne 1\) and has a basic shape ________.
A ________ function has equation \(f\left(x\right)={log}_{a}x, a>0, x>0\) and has a basic shape ________.

The image shows three scatter diagrams: linear, exponential, and logarithmic.

A Scatter Plot is a graph of plotted points that may show a relationship between the variables in a set of data.

Warm-up Example 1

Draw and interpret scatter diagrams (linear, exponential, logarithmic).

The table below shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit (Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010) . Plot this data, and determine whether the data appears to be linearly related.

Table 1
Chirps443520.433313518.53726
Temperature80.570.5576668725273.553

Plotting this data, as depicted below, suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so.

Figure 1 — Scatter plot of cricket chirps vs. air temperature

Practice Makes Perfect

Draw and interpret scatter diagrams ( linear, exponential, logarithmic).

P1

Make a scatter plot for the table below. Does it look linear? Exponential? Logarithmic?

Table 2
x123456789
y01.52.22.83.53.63.94.34.4
A blank Cartesian coordinate plane with labeled x and y axes, displaying a grid. The x-axis extends from negative to positive values, and the y-axis also extends from negative to positive values.
P2

Make a scatter plot for the table below. Does it look linear? Exponential? Logarithmic?

Table 3
x123456789
y3.35.69.115.124.440.266.2108.4180.1
A blank Cartesian coordinate plane with labeled x and y axes, featuring a grid background. The x-axis ranges from approximately -5 to 9, and the y-axis from -110 to 210.
P3

Make a scatter plot for the table below. Does it look linear? Exponential? Logarithmic?

Table 4
x123456
y35.571012.114.9
A blank Cartesian coordinate plane with labeled x and y axes, featuring a grid background. The x-axis ranges from approximately -5 to 10, and the y-axis from -10 to 20.

Objective 2: Fit a regression equation to a set of data and use the linear (or exponential) model to make predictions. (CA 4.3.4)

We can find a linear function that fits the data in the previous problem by “eyeballing” a line that seems to fit. But while estimating a line works relatively well, technology can help us find a line that fits the data as perfect as possible.

This line is called the Least Squares Regression Line or Linear Regression Model.

A regression line is a line that is closest to the data in the scatter plot, which means that such a line is a best fit for the data.

Fit a regression equation to a set of data and use the linear (or exponential) model to make predictions.

How To

Given data of input and corresponding outputs from a linear function, find the best fit line using linear regression.

  • Enter the input in List 1 (L1).
  • Enter the output in List 2 (L2).
  • On a graphing utility, select Linear Regression (LinReg).
Warm-up Example 2

Find the linear regression line using the cricket-chirp data in the example earlier in this section, and find the temperature if there are 30 chirps in 15 seconds.

Enter the input (chirps) in List 1.

  • Enter the output (temperature) in List 2.
    Table 5
    L1443520.433313518.53726
    L280.570.5576668725273.553
  • On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data, with technology we obtain the equation: T(c)=30.281+1.143c
  • To find the temperature for 30 chirps in 15 seconds we substitute 30 for x and find T:
    \(T(30)=30.281+1.143(30) =64.571\approx 64.6\,degrees\)
  • The graph of the scatter plot with the regression line of best fit is shown.

Practice Makes Perfect

Fit a regression equation to a set of data and use the linear (or exponential) model to make predictions.

P4

Gasoline consumption in the United States has been steadily increasing from 1994 to 2004.

Table 6
Year9495969798990001020304
Consumption
(billions of gallons)		113116118119123125126128131133136
  • ⓐ Determine whether the trend is linear, and if so, use your graphing utility to find a model for the data.
  • ⓑ Use the model to predict the consumption in 2008.
P5

We determined in the second practice problem, earlier in this section, that the data below has an exponential trend. Use your graphing utility to find an exponential model that fits the data the best and write your exponential model below (Hint: instead of choosing Linear Regression, choose Exponential Regression).

Table 7
x123456789
y3.35.69.115.124.440.266.2108.4180.1

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

As part of the results, your calculator will display a number known as the correlation coefficient, labeled by the variable \(r,\) or \({r}^{2}.\) (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 \({r}^{2}\) instead of \(r,\) but the closer either value is to 1, the better the regression equation approximates the data.

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, \(y=a{b}^{x}\)

Note that:

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

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

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

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 8 shows results from the study (Source: Indiana University Center for Studies of Law in Action, 2007). 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 8
BAC00.010.030.050.070.09
Relative Risk of Crashing11.031.061.382.093.54
BAC0.110.130.150.170.190.21
Relative Risk of Crashing6.4112.622.139.0565.3299.78
  • Let \(x\) represent the BAC level, and let \(y\) represent the corresponding relative risk. Use exponential regression to fit a model to these data.
  • After 6 drinks, a person weighing 160 pounds will have a BAC of about \(0.16.\) 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.

Enter the BAC values and relative-risk values into two lists, then run your calculator's exponential regression (ExpReg) to get a and b.

  • 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 2: Use the “ExpReg” command from the STAT then CALC menu to obtain the exponential model,Converting from scientific notation, we have:Notice that \({r}^{2}\approx 0.97\) 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 3:
    Figure 2 — Scatter plot of the BAC data

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

    \[y=0.58304829{(2.20720213\text{E}10)}^{x}\]

    Converting from scientific notation, we have:

    \[y=0.58304829{(\text{22,072,021,300})}^{x}\]

    Notice that \({r}^{2}\approx 0.97\) 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 3:

    Figure 3 — The BAC scatter plot with the exponential regression curve
  • Use the model to estimate the risk associated with a BAC of \(0.16.\) Substitute \(0.16\) for \(x\) in the model and solve for \(y.\) If a 160-pound person drives after having 6 drinks, they are about 26.35 times more likely to crash than if driving while sober.

    Use the model to estimate the risk associated with a BAC of \(0.16.\) Substitute \(0.16\) for \(x\) in the model and solve for \(y.\)

    \[\begin{array}{lll}y & =0.58304829{(\text{22,072,021,300})}^{x} & \text{Use the regression model found in part (a)}\text{.} \\ & =0.58304829{(\text{22,072,021,300})}^{0.16} & \text{Substitute 0}\text{.16 for }x\text{.} \\ & \approx \text{26}\text{.35} & \text{Round to the nearest hundredth}\text{.}\end{array}\]

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

Try It #1

Table 9 shows a recent graduate’s credit card balance each month after graduation.

Table 9
Month12345678
Debt ($)620.00761.88899.801039.931270.631589.041851.312154.92

ⓐ Use exponential regression to fit a model to these data.
ⓑ If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating?

  • ⓐ The exponential regression model that fits these data is \(y=522.88585984{(1.19645256)}^{x}.\)
  • ⓑ If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year.
Did you get it?
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, \(y=a+bln(x).\) For this function

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,

\[y=a+bln(x)\]

Note that

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

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

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

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 10 shows the average life expectancies, in years, of Americans from 1900–2010 (Source: Center for Disease Control and Prevention, 2013).

Table 10
Year190019101920193019401950
Life Expectancy(Years)47.350.054.159.762.968.2
Year196019701980199020002010
Life Expectancy(Years)69.770.873.775.476.878.7
  • ⓐ Let \(x\) represent time in decades starting with \(x=1\) for the year 1900, \(x=2\) for the year 1910, and so on. Let \(y\) represent the corresponding life expectancy. Use logarithmic regression to fit a model to these data.
  • ⓑ Use the model to predict the average American life expectancy for the year 2030.

Run logarithmic regression (LnReg) on the (decade, life-expectancy) pairs, then plug the target decade number into the resulting model for part (b).

  • ⓐ 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 4: Use the “LnReg” command from the STAT then CALC menu to obtain the logarithmic model, Next, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 5:
    Figure 4 — Scatter plot of the life-expectancy data

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

    \[y=42.52722583+13.85752327ln(x)\]

    Next, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 5:

    Figure 5 — The life-expectancy scatter plot with the logarithmic regression curve
  • ⓑ To predict the life expectancy of an American in the year 2030, substitute \(x=14\) for the in the model and solve for \(y:\) If life expectancy continues to increase at this pace, the average life expectancy of an American will be 79.1 by the year 2030.

    \[\begin{array}{lll}y & =42.52722583+13.85752327ln(x) & \text{Use the regression model found in part (a)}\text{.} \\ & =42.52722583+13.85752327ln(14) & \text{Substitute 14 for }x\text{.} \\ & \approx \text{79}\text{.1} & \text{Round to the nearest tenth.}\end{array}\]

    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 #2

Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. Table 11 shows the number of games sold, in thousands, from the years 2000–2010.

Table 11
Year200020012002200320042005
Number Sold (thousands)142149154155159161
Year20062007200820092010-
Number Sold (thousands)163164164166167-
  • ⓐ Let \(x\) represent time in years starting with \(x=1\) for the year 2000. Let \(y\) represent the number of games sold in thousands. Use logarithmic regression to fit a model to these data.
  • ⓑ If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand.
  • ⓐ The logarithmic regression model that fits these data is \(y=141.91242949+10.45366573ln(x)\)
  • ⓑ If sales continue at this rate, about 171,000 games will be sold in the year 2015.
Did you get it?

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:

\[y=\frac{c}{1+a{e}^{-bx}}\]

Recall that:

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

\[y=\frac{c}{1+a{e}^{-bx}}\]

Note that

  • The initial value of the model is \(\frac{c}{1+a}.\)
  • Output values for the model grow closer and closer to \(y=c\) as time increases.
How To

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

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

Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. Table 12 shows the percentage of Americans with cellular service between the years 1995 and 2012 (Source: The World Bank, 2013).

Table 12
YearAmericans with Cellular Service (%)YearAmericans with Cellular Service (%)
199512.69200462.852
199616.35200568.63
199720.29200676.64
199825.08200782.47
199930.81200885.68
200038.75200989.14
200145.00201091.86
200249.16201195.28
200355.15201298.17
  • ⓐ Let \(x\) represent time in years starting with \(x=0\) for the year 1995. Let \(y\) represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data.
  • ⓑ Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent.
  • ⓒ Discuss the value returned for the upper limit, \(c.\) What does this tell you about the model? What would the limiting value be if the model were exact?

Run logistic regression on the (year, percentage) pairs to get a, b, c, then compare the returned c to the true maximum possible percentage.

  • ⓐ 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 6: Use the “Logistic” command from the STAT then CALC menu to obtain the logistic model,Next, graph the model in the same window as shown in Figure 7 the scatterplot to verify it is a good fit:
    Figure 6 — Scatter plot of the cellular-service data

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

    \[y=\frac{105.7379526}{1+6.88328979{e}^{-0.2595440013x}}\]

    Next, graph the model in the same window as shown in Figure 7 the scatterplot to verify it is a good fit:

    Figure 7 — The cellular-service scatter plot with the logistic regression curve
  • ⓑ To approximate the percentage of Americans with cellular service in the year 2013, substitute \(x=18\) for the in the model and solve for \(y:\)According to the model, about 99.3% of Americans had cellular service in 2013.

    To approximate the percentage of Americans with cellular service in the year 2013, substitute \(x=18\) for the in the model and solve for \(y:\)

    \[\begin{array}{lll}y & =\frac{105.7379526}{1+6.88328979{e}^{-0.2595440013x}} & \text{Use the regression model found in part (a)}. \\ & =\frac{105.7379526}{1+6.88328979{e}^{-0.2595440013(18)}} & \text{Substitute 18 for }x. \\ & \approx \text{99}\text{.3 } & \text{Round to the nearest tenth}\end{array}\]

    According to the model, about 99.3% of Americans had cellular service in 2013.

  • ⓒ 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 \(c=100\) 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!

    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 \(c=100\) 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 #3

Table 13 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.

Table 13
YearSeal Population (Thousands)YearSeal Population (Thousands)
19973.493200519.590
19985.282200621.955
19996.357200722.862
20009.201200823.869
200111.224200924.243
200212.964201024.344
200316.226201124.919
200418.137201225.108
  • ⓐ Let \(x\) represent time in years starting with \(x=0\) for the year 1997. Let \(y\) represent the number of seals in thousands. Use logistic regression to fit a model to these data.
  • ⓑ Use the model to predict the seal population for the year 2020.
  • ⓒ To the nearest whole number, what is the limiting value of this model?
  • ⓐ The logistic regression model that fits these data is \(y=\frac{25.65665979}{1+6.113686306{e}^{-0.3852149008x}}.\)
  • ⓑ If the population continues to grow at this rate, there will be about \(\text{25,634}\) seals in 2020.
  • ⓒ To the nearest whole number, the carrying capacity is 25,657.
Did you get it?
Media

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

Key Concepts

Section Exercises

Verbal

1

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

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.

2

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

3

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

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.

4

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

5

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

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 8 through Figure 12. Answer using the letter beneath the matching graph.

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.
Figure 12
6

\(y=10.209{e}^{-0.294x}\)

7

\(y=5.598-1.912ln(x)\)

C

8

\(y=2.104{(1.479)}^{x}\)

9

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

B

10

\(y=\frac{14.005}{1+2.79{e}^{-0.812x}}\)

Numeric

11

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

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

12

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

13

A logarithmic model is given by the equation \(h(p)=67.682-5.792ln(p).\) To the nearest hundredth, for what value of \(p\) does \(h(p)=62?\)

\(p\approx 2.67\)

14

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

15

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

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

Technology

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

16

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

17

What was the initial population of koi?

\(4\) koi

18

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

19

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

about \(6.8\) months.

20

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

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

21

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

A two-dimensional line graph with an x-axis ranging from 0 to 20 and a y-axis ranging from 0 to 600. A dark blue, S-shaped curve begins near the origin, approximately at (1, 15), and ascends through the graph, passing roughly through points (5, 80), (10, 300), and (15, 500), before ending with an arrow around (17.5, 550) towards the top right.
22

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

23

How many wolves will the habitat have after \(3\) years?

About 38 wolves

24

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

25

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

About 8.7 years

For the following exercises, refer to Table 14.

Table 14
x123456
f(x)112514952310329446506361
26

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

27

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

\(f(x)=776.682{(1.426)}^{x}\)

28

Write the exponential function as an exponential equation with base \(e.\)

29

Graph the exponential equation on the scatter diagram.

A graph illustrating exponential growth, showing an orange curve with six blue data points plotted. As the x-value increases, the y-value rapidly increases, indicating a steep upward trend.
30

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

For the following exercises, refer to Table 15.

Table 15
x123456
f(x)555383307210158122
31

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

A scatter plot on a grid shows six blue data points with x-values from 1 to 6 and corresponding y-values decreasing from 550 to 120, illustrating a negative correlation.
32

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

33

Write the exponential function as an exponential equation with base \(e.\)

\(f(x)={731.92e}^{-0.3038x}\)

34

Graph the exponential equation on the scatter diagram.

35

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

When \(f(x)=250,\,x\approx 3.6\)

For the following exercises, refer to Table 16.

Table 16
x123456
f(x)5.16.37.37.78.18.6
36

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

37

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

\(y=5.063+1.934\text{log}(x)\)

38

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

39

Graph the logarithmic equation on the scatter diagram.

An image showing an orange curve passing through six blue data points on a grid with x-axis from 0 to 7 and y-axis from 0 to 10, indicating an increasing trend.
40

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

For the following exercises, refer to Table 17.

Table 17
x12345678
f(x)7.565.24.33.93.43.12.9
41

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

A scatter plot shows a series of data points illustrating a decreasing trend. As the x-value increases from 1 to 8, the corresponding y-value generally decreases from approximately 7.5 to 2.9, indicating a negative correlation.
42

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

43

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

When \(f(10)\approx 2.3\)

44

Graph the logarithmic equation on the scatter diagram.

45

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

When \(f(x)=8,\,x\approx 0.82\)

For the following exercises, refer to Table 18.

Table 18
x12345678910
f(x)8.712.315.418.520.722.523.32424.624.8
46

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

47

Use the LOGISTIC regression option to find a logistic growth model of the form \(y=\frac{c}{1+a{e}^{-bx}}\) that best fits the data in the table.

\(f(x)=\frac{25.081}{1+3.182{e}^{-0.545x}}\)

48

Graph the logistic equation on the scatter diagram.

49

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

About 25

50

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

For the following exercises, refer to Table 19.

Table 19
\(x\)02457810111517
\(f(x)\)1228.652.870.399.9112.5125.8127.9135.1135.9
51

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

A scatter plot shows ten data points on a grid with an x-axis from 0 to 18 and a y-axis from 0 to 140. The points exhibit a strong increasing trend.
52

Use the LOGISTIC regression option to find a logistic growth model of the form \(y=\frac{c}{1+a{e}^{-bx}}\) that best fits the data in the table.

53

Graph the logistic equation on the scatter diagram.

A graph with an x-axis labeled from 0 to 18 and a y-axis labeled from 0 to 140. An orange S-shaped curve is plotted, starting near (0, 11), increasing steeply through points such as (2, 29), (4, 53), (5, 70), (7, 100), (8, 112), and (10, 126), and then flattening out as it approaches a maximum y-value around 135-136, with data points at (11, 128) and (15, 134). The curve suggests logistic growth.
54

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

55

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

When \(f(x)=68,\,x\approx 4.9\)

Extensions

56

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

57

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

\(f(x)=1.034341{(1.281204)}^{x}\) ; \(g(x)=4.035510\) ; the regression curves are symmetrical about \(y=x\) , so it appears that they are inverse functions.

58

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

59

Find the inverse function \({f}^{-1}(x)\) for the logistic function \(f(x)=\frac{c}{1+a{e}^{-bx}}.\) Show all steps.

\({f}^{-1}(x)=\frac{\text{ln}(a)-\text{ln}(\frac{c}{x}-1)}{b}\)

60

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

Chapter Review Exercises

Exponential Functions

1

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

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

2

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

3

Find an exponential equation that passes through the points \(\text{(2, 2}\text{.25)}\) and \((5,60.75).\)

\(y=0.25{(3)}^{x}\)

4

Determine whether Table 20 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 20
x1234
f(x)30.90.270.081
5

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

\(\$42,888.18\)

6

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 \(7.5\%\) APR, compounded daily, in order to reach her goal in \(3\) years?

7

Does the equation \(y=2.294{e}^{-0.654t}\) represent continuous growth, continuous decay, or neither? Explain.

continuous decay; the growth rate is negative.

8

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

Graphs of Exponential Functions

9

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

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)
10

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

11

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

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

12

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

Graph of f(x)=2^x
Figure 13

Logarithmic Functions

13

Rewrite \({log}_{17}(4913)=x\) as an equivalent exponential equation.

\({17}^{x}=4913\)

14

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

15

Rewrite \({a}^{-\,\frac{2}{5}}=b\) as an equivalent logarithmic equation.

\({log}_{a}b=-\frac{2}{5}\)

16

Rewrite \({e}^{-3.5}=h\) as an equivalent logarithmic equation.

17

Solve for x if \({log}_{64}(x)=\frac{1}{3}\) by converting the logarithmic equation \({log}_{64}(x)=\frac{1}{3}\) to exponential form.

\(x={64}^{\frac{1}{3}}=4\)

18

Evaluate \({log}_{5}(\frac{1}{125})\) without using a calculator.

19

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

\(log(\text{0}\text{.000001})=-6\)

20

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

21

Evaluate \(ln({e}^{-0.8648})\) without using a calculator.

\(ln({e}^{-0.8648})=-0.8648\)

22

Evaluate \(ln(\sqrt[3]{18})\) using a calculator. Round to the nearest thousandth.

Graphs of Logarithmic Functions

23

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


Graph of g(x)=log(7x+21)-4.
24

Graph the function \(h(x)=2ln(9-3x)+1.\)

25

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

Domain: \(x>-5;\) Vertical asymptote: \(x=-5;\) End behavior: as \(x\to -{5}^{+},f(x)\to -\infty\) and as \(x\to \infty ,f(x)\to \infty .\)

Logarithmic Properties

26

Rewrite \(ln(7r\cdot 11st)\) in expanded form.

27

Rewrite \({log}_{8}(x)+{log}_{8}(5)+{log}_{8}(y)+{log}_{8}(13)\) in compact form.

\({\text{log}}_{8}(65xy)\)

28

Rewrite \({log}_{m}(\frac{67}{83})\) in expanded form.

29

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

\(ln(\frac{z}{xy})\)

30

Rewrite \(ln(\frac{1}{{x}^{5}})\) as a product.

31

Rewrite \(-{log}_{y}(\frac{1}{12})\) as a single logarithm.

\({\text{log}}_{y}(12)\)

32

Use properties of logarithms to expand \(log(\frac{{r}^{2}{s}^{11}}{{t}^{14}}).\)

33

Use properties of logarithms to expand \(ln(2b\sqrt{\frac{b+1}{b-1}}).\)

\(ln(2)+ln(b)+\frac{ln(b+1)-ln(b-1)}{2}\)

34

Condense the expression \(5ln(b)+ln(c)+\frac{ln(4-a)}{2}\) to a single logarithm.

35

Condense the expression \(3{log}_{7}v+6{log}_{7}w-\frac{{log}_{7}u}{3}\) to a single logarithm.

\({log}_{7}(\frac{{v}^{3}{w}^{6}}{\sqrt[3]{u}})\)

36

Rewrite \({log}_{3}(12.75)\) to base \(e.\)

37

Rewrite \({5}^{12x-17}=125\) as a logarithm. Then apply the change of base formula to solve for \(x\) using the common log. Round to the nearest thousandth.

\(x=\frac{\frac{log(125)}{log(5)}+17}{12}=\frac{5}{3}\)

Exponential and Logarithmic Equations

38

Solve \({216}^{3x}\cdot {216}^{x}={36}^{3x+2}\) by rewriting each side with a common base.

39

Solve \(\frac{125}{{(\frac{1}{625})}^{-x-3}}={5}^{3}\) by rewriting each side with a common base.

\(x=-3\)

40

Use logarithms to find the exact solution for \(7\cdot {17}^{-9x}-7=49.\) If there is no solution, write no solution.

41

Use logarithms to find the exact solution for \(3{e}^{6n-2}+1=-60.\) If there is no solution, write no solution.

no solution

42

Find the exact solution for \(5{e}^{3x}-4=6\) . If there is no solution, write no solution.

43

Find the exact solution for \(2{e}^{5x-2}-9=-56.\) If there is no solution, write no solution.

no solution

44

Find the exact solution for \({5}^{2x-3}={7}^{x+1}.\) If there is no solution, write no solution.

45

Find the exact solution for \({e}^{2x}-{e}^{x}-110=0.\) If there is no solution, write no solution.

\(x=ln(11)\)

46

Use the definition of a logarithm to solve. \(-5{log}_{7}(10n)=5.\)

47

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

\(a={e}^{4}-3\)

48

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

49

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

\(x=\pm \frac{9}{5}\)

50

The formula for measuring sound intensity in decibels \(D\) is defined by the equation \(D=10log(\frac{I}{{I}_{0}}),\) where \(I\) is the intensity of the sound in watts per square meter and \({I}_{0}={10}^{-12}\) 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 \(6.3\cdot {10}^{-3}\) watts per square meter?

51

The population of a city is modeled by the equation \(P(t)=256,114{e}^{0.25t}\) where \(t\) 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?

about \(5.45\) years

52

Find the inverse function \({f}^{-1}\) for the exponential function \(f(x)=2\cdot {e}^{x+1}-5.\)

53

Find the inverse function \({f}^{-1}\) for the logarithmic function \(f(x)=0.25\cdot {log}_{2}({x}^{3}+1).\)

\({f}^{-1}(x)=\sqrt[3]{{2}^{4x}-1}\)

Exponential and Logarithmic Models

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

54

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

55

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

\(f(t)=300{(0.83)}^{t};\)
\(f(24)\approx 3.43\,\,g\)

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

56

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

57

How many minutes will it take the soup to cool to \(\text{85°F?}\)

about \(45\) minutes

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

58

How many people started the rumor?

59

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

about \(8.5\) days

60

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.

61
Table 21
xf(x)
13.05
24.42
36.4
49.28
513.46
619.52
728.3
841.04
959.5
1086.28

exponential

Graph of the table’s values.
62
Table 22
xf(x)
0.518.05
117
315.33
514.55
714.04
1013.5
1213.22
1313.1
1512.88
1712.69
2012.45
63

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

\(y=4{(0.2)}^{x};\) \(y=4{e}^{\text{-1.609438}x}\)

Fitting Exponential Models to Data

64

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

65

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

about \(7.2\) 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.

66
Table 23
xf(x)
1409.4
2260.7
3170.4
4110.6
574
644.7
732.4
819.5
912.7
108.1
67
Table 24
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

logarithmic; \(y=16.68718-9.71860ln(x)\)

Graph of the table’s values.
68
Table 25
xf(x)
09
222.6
444.2
562.1
796.9
8113.4
10133.4
11137.6
15148.4
17149.3

Practice Test

1

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

About 13 dolphins.

2

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

3

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 \(6.25\%\) APR, compounding daily, in order to reach his goal in \(4\) years?

\(\$1,947\)

4

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

5

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

y-intercept: \((0,\,\text{5})\)

Graph of f(-x)=5(0.5)^-x in blue and f(x)=5(0.5)^x in orange.
6

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

Graph of f(x)= (1/2)^x.
7

Rewrite \({log}_{8.5}(614.125)=a\) as an equivalent exponential equation.

\({8.5}^{a}=614.125\)

8

Rewrite \({e}^{\frac{1}{2}}=m\) as an equivalent logarithmic equation.

9

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

\(x={(\frac{1}{7})}^{2}=\frac{1}{49}\)

10

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

11

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

\(ln(0.716)\approx -0.334\)

12

Graph the function \(g(x)=log(12-6x)+3.\)

13

State the domain, vertical asymptote, and end behavior of the function \(f(x)={log}_{5}(39-13x)+7.\)

Domain: \(x<3;\) Vertical asymptote: \(x=3;\) End behavior: \(x\to {3}^{-},f(x)\to -\infty\) and \(x\to -\infty ,f(x)\to \infty\)

14

Rewrite \(log(17a\cdot 2b)\) as a sum.

15

Rewrite \({log}_{t}(96)-{log}_{t}(8)\) in compact form.

\({log}_{t}(12)\)

16

Rewrite \({log}_{8}({a}^{\frac{1}{b}})\) as a product.

17

Use properties of logarithm to expand \(ln({y}^{3}{z}^{2}\cdot \sqrt[3]{x-4}).\)

\(3\,ln(y)+2ln(z)+\frac{ln(x-4)}{3}\)

18

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

19

Rewrite \({16}^{3x-5}=1000\) as a logarithm. Then apply the change of base formula to solve for \(x\) using the natural log. Round to the nearest thousandth.

\(x=\frac{\frac{ln(1000)}{ln(16)}+5}{3}\approx 2.497\)

20

Solve \({(\frac{1}{81})}^{x}\cdot \frac{1}{243}={(\frac{1}{9})}^{-3x-1}\) by rewriting each side with a common base.

21

Use logarithms to find the exact solution for \(-9{e}^{10a-8}-5=-41\) . If there is no solution, write no solution.

\(a=\frac{ln(4)+8}{10}\)

22

Find the exact solution for \(10{e}^{4x+2}+5=56.\) If there is no solution, write no solution.

23

Find the exact solution for \(-5{e}^{-4x-1}-4=64.\) If there is no solution, write no solution.

no solution

24

Find the exact solution for \({2}^{x-3}={6}^{2x-1}.\) If there is no solution, write no solution.

25

Find the exact solution for \({e}^{2x}-{e}^{x}-72=0.\) If there is no solution, write no solution.

\(x=ln(9)\)

26

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

27

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

\(x=\pm \frac{3\sqrt{3}}{2}\)

28

The formula for measuring sound intensity in decibels \(D\) is defined by the equation \(D=10log(\frac{I}{{I}_{0}}),\) where \(I\) is the intensity of the sound in watts per square meter and \({I}_{0}={10}^{-12}\) 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 \(4.7\cdot {10}^{-1}\) watts per square meter?

29

A radiation safety officer is working with \(112\) grams of a radioactive substance. After \(17\) days, the sample has decayed to \(80\) 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?

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

30

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

31

A bottle of soda with a temperature of \(\text{71°}\) Fahrenheit was taken off a shelf and placed in a refrigerator with an internal temperature of \(\text{35° F}\text{.}\) After ten minutes, the internal temperature of the soda was \(\text{63° F}\text{.}\) 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?

\(T(t)=36{e}^{-0.025131t}+35;T(60)\approx {43}^{\text{o}}\text{F}\)

32

The population of a wildlife habitat is modeled by the equation \(P(t)=\frac{360}{1+6.2{e}^{-0.35t}},\) where \(t\) 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?

33

Enter the data from Table 26 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 26
xf(x)
13
28.55
311.79
414.09
515.88
617.33
718.57
819.64
920.58
1021.42

logarithmic

Graph of the table’s values.
34

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

35
Table 27
xf(x)
120
221.6
329.2
436.4
546.6
655.7
772.6
887.1
9107.2
10138.1

exponential; \(y=15.10062{(1.24621)}^{x}\)

Graph of the table’s values.
36
Table 28
xf(x)
313.98
417.84
520.01
622.7
724.1
826.15
927.37
1028.38
1129.97
1231.07
1331.43
37
Table 29
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

logistic; \(y=\frac{18.41659}{1+7.54644{e}^{-0.68375x}}\)

Graph of the table’s values.