Fitting Exponential Models to Data

Fitting Exponential Models to Data

• 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)

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

Example

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

Try it.

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.

Chirps	44	35	20.4	33	31	35	18.5	37	26
Temperature	80.5	70.5	57	66	68	72	52	73.5	53

Solution

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.

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

Try it.

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

x	1	2	3	4	5	6	7	8	9
y	0	1.5	2.2	2.8	3.5	3.6	3.9	4.3	4.4

Try it.

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

x	1	2	3	4	5	6	7	8	9
y	3.3	5.6	9.1	15.1	24.4	40.2	66.2	108.4	180.1

Try it.

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

x	1	2	3	4	5	6
y	3	5.5	7	10	12.1	14.9

Condensed — the full section is in OpenStax College Algebra 2e.

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 \(a>0):\)

• \(b\) must be greater than zero and not equal to one.
• The initial value of the model is \(y=a.\)
  • If \(b>1,\) the function models exponential growth. As \(x\) increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound.
  • If \(0exponential decay. As \(x\) 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 \(r,\) 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.

Condensed — the full section is in OpenStax College Algebra 2e.

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, \(y=a+b\ln (x).\) For this function

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

Condensed — the full section is in OpenStax College Algebra 2e.

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:

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

Condensed — the full section is in OpenStax College Algebra 2e.

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 \(y=a{b}^{x}\) to a set of data points. See .
• 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 \(y=a+b\ln (x)\) to a set of data points. See .
• 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 \(y=\frac{c}{1+a{e}^{-bx}}\) to a set of data points. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. 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.

   Chirps	44	35	20.4	33	31	35	18.5	37	26
   Temperature	80.5	70.5	57	66	68	72	52	73.5	53

   Jawaabta muuji

   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.

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

   x	1	2	3	4	5	6	7	8	9
   y	0	1.5	2.2	2.8	3.5	3.6	3.9	4.3	4.4

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

   x	1	2	3	4	5	6	7	8	9
   y	3.3	5.6	9.1	15.1	24.4	40.2	66.2	108.4	180.1

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

   x	1	2	3	4	5	6
   y	3	5.5	7	10	12.1	14.9

5. 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.

   Jawaabta muuji

   Enter the input (chirps) in List 1.

   1. Enter the output (temperature) in List 2.
      

      L1	44	35	20.4	33	31	35	18.5	37	26
      L2	80.5	70.5	57	66	68	72	52	73.5	53

   2. 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
   3. To find the temperature for 30 chirps in 15 seconds we substitute 30 for x and find T:
      \(T(30)=30.281+1.143(30)=64.571\approx 64.6\ degrees\)
   4. The graph of the scatter plot with the regression line of best fit is shown.

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

   Year	94	95	96	97	98	99	00	01	02	03	04
   Consumption (billions of gallons)	113	116	118	119	123	125	126	128	131	133	136

   • ⓐ 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.

7. 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).

   x	1	2	3	4	5	6	7	8	9
   y	3.3	5.6	9.1	15.1	24.4	40.2	66.2	108.4	180.1

8. 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. 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.

   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 \(x\) represent the BAC level, and let \(y\) 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 \(0.16.\) 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.
   Jawaabta muuji

   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 :

      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 :

   2. 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.

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

   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

   ⓐ 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?

   Jawaabta muuji

   1. ⓐ The exponential regression model that fits these data is \(y=522.88585984{(1.19645256)}^{x}.\)
   2. ⓑ If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year.

10. 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.

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

    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 \(x\) 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.
    2. ⓑ Use the model to predict the average American life expectancy for the year 2030.
    Jawaabta muuji

    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 :

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

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

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

    2. ⓑ To predict the life expectancy of an American in the year 2030, substitute \(x=14\) for the in the model and solve for \(y:\) \[\begin{array}{lll}y & =42.52722583+13.85752327\ln (x) & \text{Use the regression model found in part (a)}\text{.} \\ & =42.52722583+13.85752327\ln (14) & \text{Substitute 14 for }x\text{.} \\ & \approx \text{79}\text{.1} & \text{Round to the nearest tenth.}\end{array}\]

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

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

    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	-

    • ⓐ 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.
    Jawaabta muuji

    1. ⓐ The logarithmic regression model that fits these data is \(y=141.91242949+10.45366573\ln (x)\)
    2. ⓑ If sales continue at this rate, about 171,000 games will be sold in the year 2015.

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

    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

    • ⓐ 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?
    Jawaabta muuji

    • ⓐ 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 :

      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 the scatterplot to verify it is a good fit:

    • ⓑ

      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!

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

    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

    • ⓐ 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?
    Jawaabta muuji

    1. ⓐ The logistic regression model that fits these data is \(y=\frac{25.65665979}{1+6.113686306{e}^{-0.3852149008x}}.\)
    2. ⓑ If the population continues to grow at this rate, there will be about \(\text{25,634}\) seals in 2020.
    3. ⓒ To the nearest whole number, the carrying capacity is 25,657.

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

    Jawaabta muuji

    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.

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

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

    Jawaabta muuji

    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.

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

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

    Jawaabta muuji

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

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

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

    Jawaabta muuji

    C

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

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

    Jawaabta muuji

    B

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

24. 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?

    Jawaabta muuji

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

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

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

    Jawaabta muuji

    \(p\approx 2.67\)

27. 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 \(P(t)=45?\)

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

    Jawaabta muuji

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

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

30. What was the initial population of koi?

    Jawaabta muuji

    \(4\) koi

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

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

    Jawaabta muuji

    about \(6.8\) months.

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

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

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

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

    Jawaabta muuji

    About 38 wolves

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

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

    Jawaabta muuji

    About 8.7 years

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

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

    Jawaabta muuji

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

Symbols used here

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

i

imaginary unit
i² = −1.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Fitting Exponential Models to Data

1. Build an exponential model from data.
2. Build a logarithmic model from data.
3. Build a logistic model from data.
4. Draw and interpret scatter diagrams (linear, exponential, logarithmic). (CA 4.3.1)
5. Fit a regression equation to a set of data and use the linear (or exponential) model to make predictions. (CA 4.3.4)
6. Enter the input in List 1 (L1).
7. Enter the output in List 2 (L2).
8. On a graphing utility, select Linear Regression (LinReg).

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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⌨ Type ✎ Draw

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Xalin

x²√÷×π( )=∫d/dxlim[ ]sinlogeˣ

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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