3.5: Exponential and Logarithmic Models and Equations
Learning objectives
 Analyze data to determine if it can be represented by an exponential or logarithmic model.
 Use a graphing calculator to find an exponential or logarithmic model, and use a model to answer questions about a situation.
 Solve exponential and logarithmic equations using properties of exponents and logarithms.
 Find approximate solutions to equations using a graphing calculator.
Introduction
So far in this chapter we have evaluated exponential and logarithm expressions, and we have graphed exponential and logarithmic functions. In this lesson you will extend what you have learned in two ways. First, we will introduce the idea of modeling real phenomena with logarithmic and exponential functions. Second, we will solve logarithmic and exponential equations. While you have already solved some equations in previous lessons, now you will be able to solve more complicated equations. This lesson will provide you with further tools for the applications of logarithmic and exponential functions that will be the focus of the remainder of the chapter.
Exponential Models
Consider the following example: the population of a small town was 2,000 in the year 1950. The population increased over time, as shown by the values in the table:
Year (1950 = 0)  Population 

0  2000 
5  2980 
10  4450 
20  9900 
30  22,000 
40  50,000 
If you plot these data points, you will see that the growth pattern is nonlinear:
In many situations, population growth can be modeled with an exponential function (this is because population grows as a percentage of the current population, i.e. 8% per year). In lesson 7, you will learn how to create such models using information from a given situation. Here, we will focus on creating models using data and a graphing calculator.
The population data from the example above can be modeled with an exponential function, but the function is not unique. That is, there is more than one way to write a function to model this data. In the steps below you will see how to use a graphing calculator to find a function of the form y = a(b^{x}) that fits the data in the table.
 Technology Note
 Using a TI83/84 graphing calculator to find an exponential function that best fits a set of data

1. Entering the data
 a. Data must be entered into “lists”. The calculator has six named lists, L1, L2, ... L6. We will enter the x values in L1 and the y values in L2. One way to do this is shown below:
 Press <TI font_2nd> [{] and then enter the numbers separated by commas, and close by pressing the following: <TI font_2nd> [{]<TI font_STO> <TI font_2nd> [L1].
 The top three lines of the figure below show the entry into list L1, followed by the entry of the y values into list L2.
 Now press <TI font_STAT>, and move to the right to the CALC menu. Scroll down to option 10, ExpReg. Press <TI font_ENTER>, and you will return to the home screen. You should see ExpReg on the screen. As long as the numbers are in L1 and L2, the calculator will proceed to find an exponential function to fit the data you listed in List L1 and List L2. You should see on the home screen the values for a and b in the exponential function (See figure below). Therefore the function y = 1992.7(1.0837)^{x} is an approximate model for the data.
 2. Plotting the data and the equation

To view plots of the data points and the equation on the same screen, do the following.
 a. First, press <TI font_Y=> and clear any equations.

You can type in the equation above, or to get the equation from the calculator, do the following:

b. Enter the above roundedoff equation in Y1, or use the following procedure to get the full equation from the calculator: put the cursor in Y1, press <TI font_VARS>, 5, EQ, and 1. This should place the equation in Y1 (see figure below).

c. Now press <TI font_2nd>[STAT PLOT] and complete the items as shown in the figure below.

d. Now set your window. (Hint: use the range of the data to choose the window – the figure below shows our choices.)

e. Press <TI font GRAPH> and you will to see the function and the data points as shown in the figure below.

b. Enter the above roundedoff equation in Y1, or use the following procedure to get the full equation from the calculator: put the cursor in Y1, press <TI font_VARS>, 5, EQ, and 1. This should place the equation in Y1 (see figure below).
 3. Comparing the real data with the modeled results

It looks as if the data points lie on the function. However, using the TRACE function you can determine how close the modeled points are to the real data. Press <TI font_TRACE> to enter the TRACE mode. Then press the right arrow to move from one data point to another. Do this until you land on the point with value Y=22000. To see the corresponding modeled value, press the up or down arrow. See the figure below. The modeled value is approximately 22197, which is quite close to the actual data. You can verify any of the other data points using the same method.
Now that we have the equation y = 1992.7(1.0837)^{x} to model the situation, we can estimate the population for any years that were not in the original data set. If we choose x values between 0 and 40, it is called interpolation. If we choose other x values outside of this domain, it is called extrapolation. Interpolation is, in a sense, a safer way of estimating population, because it is within the data points that we have, and does not require that we think about the end behavior of the function. For example, if we extrapolate to the year 1930, this means x = 20. The function value is 399. However, if the town was founded in 1940, then this data value does not make any sense. Similarly, if we extrapolate to the year 2000, we have x = 50. The function value is 110,711. However, if the town’s pattern of population growth shifted (perhaps due to some economic change), this estimation could be highly inaccurate. As noted above, you will study exponential growth, as well as other exponential models, in the next two lessons. Now we turn to logarithmic models.
Logarithmic Models
Consider another example of population growth:
Table 2
Year  Population 

1  2000 
5  4200 
10  6500 
20  8800 
30  10500 
40  12500 
If we plot this data, we see that the growth is not quite linear, and it is not exponential either.
Just as we found an exponential model in the previous example, here we can find a logarithmic function to model this data. First enter the data in the table in L1 and L2. Then press STAT to get to the CALC menu. This time choose option 9. You should get the function y = 930.4954615 + 2780.218173 ln x. If you view the graph and the data points together, as described in the Technology Note above, you will see that the graph of the function does not touch the data points, but models the general trend of the data.
Note about technology: you can also do this using an Excel spreadsheet. Enter the data in a worksheet, and create a scatterplot by inserting a chart. After you create the chart, from the chart menu, choose “add trendline.” You will then be able to choose the type of function. Note that if you want to use a logarithmic function, the domain of your data set must be positive numbers. The chart menu will actually not allow you to choose a logarithmic trendline if your data include zero or negative x values. See below:
Solve Exponential Equations
Given an exponential model of some phenomena, such as population growth, you may want to determine a particular input value that would produce a given function value. Let’s say that a function P(x) = 2000(1.05)^{x} models the population growth for a town. What if we want to know when the population reaches 20,000?
To answer this equation, we must solve the equation 2000(1.05)^{x} = 20,000. We can solve this equation by isolating the power (1.05)^{x} and then using one of the log properties:
\begin{align*}2000(1.05)^x = 20,000\end{align*} 
Divide both sides of the equation by \begin{align*}2000\end{align*} 

\begin{align*}(1.05)^x = 10\end{align*} 
Take the common log of both sides 
\begin{align*}log(1.05)^x = log 10\end{align*} 
Use the power property of logs 
\begin{align*}x log (1.05) = log 10\end{align*} 
Evaluate \begin{align*}log 10\end{align*} 
\begin{align*}x log (1.05) = 1\end{align*} 

\begin{align*}x = \frac{1} {log(1.05)} \approx 47\end{align*} 
Divide both sides by \begin{align*}log(1.05)\end{align*}
Use a calculator to estimate \begin{align*}log(1.05)\end{align*} 
We can use these same techniques to solve any exponential equation.
Example 1: Solve each exponential equation
a. 2^{x} + 7 = 19  b. 3^{5x  1} = 16 

Solution:

a. \begin{align*}2^x + 7 = 19\end{align*}
2x+7=19


\begin{align*}7 7\end{align*}
−7−7

\begin{align*}7 7\end{align*}


\begin{align*}2^x = 12\end{align*}
2x=12

\begin{align*}2^x = 12\end{align*}


\begin{align*}log 2^x = log 12\end{align*}
log2x=log12

\begin{align*}log 2^x = log 12\end{align*}


\begin{align*}x log 2 = log 12\end{align*}
xlog2=log12

\begin{align*}x log 2 = log 12\end{align*}


\begin{align*}x = \frac{log 12} {log 2} \approx 3.58\end{align*}
x=log12log2≈3.58

\begin{align*}x = \frac{log 12} {log 2} \approx 3.58\end{align*}
b. \begin{align*}3^{5x  1} = 16\end{align*}


\begin{align*}log 3^{5x  1} = log 16\end{align*}
log35x−1=log16

\begin{align*}log 3^{5x  1} = log 16\end{align*}


\begin{align*}(5x  1) log 3 = log 16\end{align*}
(5x−1)log3=log16

\begin{align*}(5x  1) log 3 = log 16\end{align*}

 \begin{align*}5x  1 = \frac{log 16} {log 3}\end{align*}

 \begin{align*}5x = \frac{log 16} {log 3} + 1\end{align*}

 \begin{align*}x = \frac{\frac{log 16} {log 3} + 1} {5}\end{align*}

 \begin{align*}x \approx 0.705\end{align*}
Solve Logarithmic Equations
In the previous lesson we solved two forms of log equations. Now we can solve more complicated equations, using our knowledge of log properties. For example, consider the equation log_{2} (x) + log_{2} (x  2) = 3. We can solve this equation using a log property.
log_{2} (x) + log_{2} (x  2) = 3  

log_{2} (x(x  2)) = 3  log_{b} x + log_{b} y = log_{b}(xy) 
log_{2} (x^{2}  2x) = 3 \begin{align*}\Rightarrow\end{align*}  write the equation in exponential form. 
2^{3} = x^{2}  2x  
x^{2}  2x  8 = 0  Solve the resulting quadratic 
(x  4) (x + 2) = 0  
x = 2, 4 
The resulting quadratic has two solutions. However, only x = 4 is a solution to our original equataion, as log_{2}(2) is undefined. We refer to x = 2 as an extraneous solution.
Example 2: Solve each equation
a. log (x + 2) + log 3 = 2  b. ln (x + 2)  ln (x) = 1 

Solution:
a. log (x + 2) + log 3 = 2

log (3(x + 2)) = 2 log_{b} x + log_{b} y = log_{b} (xy) log (3x + 6) = 2 Simplify the expression 3(x+2) 10^{2} = 3x + 6 Write the log expression in exponential form 100 = 3x + 6 3x = 94 Solve the linear equation x = 94/3
b. ln (x + 2)  ln (x) = 1

\begin{align*}ln \left (\frac{x + 2} {x}\right ) = 1\end{align*} \begin{align*}log_b x  log_b y = log_b \left (\frac{x} {y}\right )\end{align*} \begin{align*}e^1 = \frac{x + 2} {x}\end{align*} Write the log expression in exponential form. \begin{align*}ex = x + 2\end{align*} Multiply both sides by \begin{align*}x\end{align*}. \begin{align*}ex  x = 2\end{align*} Factor out \begin{align*}x\end{align*}. \begin{align*}x (e  1) = 2\end{align*} Isolate \begin{align*}x\end{align*}. \begin{align*}x = \frac{2} {e  1}\end{align*}
The solution above is an exact solution. If we want a decimal approximation, we can use a calculator to find that x ≈ 1.16. We can also use a graphing calculator to find an approximate solution, as we did in lesson 2 with exponential equations. Consider again the equation ln (x + 2)  ln (x) = 1. We can solve this equation by solving a system:
\begin{align*}\begin{cases} y = ln (x + 2)  ln (x)\\ y = 1\\ \end{cases} \end{align*}
If you graph the system on your graphing calculator, as we did in lesson 2, you should see that the curve and the horizontal line intersection at one point. Using the INTERSECT function on the CALC menu (press <TI font_2nd>[CALC]), you should find that the x coordinate of the intersection point is approximately 1.16. This method will allow you to find approximate solutions for more complicated log equations.
Example 3. Use a graphing calculator to solve each equation:
a. log(5  x) + 1 = log x  b. log_{2}(3x + 8) + 1 = log_{3} (10  x) 

Solution:
a. log(5  x) + 1 = log x
 The graphs of y = log (5  x) + 1 and y = log x intersect at x ≈ 4.5454545
 Therefore the solution of the equation is x ≈ 4.54.
b. log_{2} (3x + 8) + 1 = log_{3} (10  x)
 First, in order to graph the equations, you must rewrite them in terms of a common log or a natural log. The resulting equations are: \begin{align*}y = \frac{log(3x + 8)}{log 2} + 1 \end{align*} and \begin{align*}y = \frac{log(10  x)}{log 3}\end{align*}. The graphs of these functions intersect at x ≈ 1.87. This value is the approximate solution to the equation.
Lesson Summary
This lesson has introduced the idea of modeling a situation using an exponential or logarithmic function. When a population or other quantity has a steep increase over time, it may be modeled with an exponential function. When a population has a steep increase, but then slower growth, it may be modeled with a logarithmic function. (In a later lesson you will learn about a third option.) We have also examined techniques of solving exponential and logarithmic equations, based on our knowledge of properties of logarithms. The key property to remember is the power property:

 log_{b} x^{n} = n log_{b} x
Using this property allows us to turn an exponential function into a linear function, which we can then solve in order to solve the original exponential function.
In the remaining lessons in this chapter, you will learn about several different real phenomena that are modeled with exponential and logarithmic equations. In these lessons you will also use the techniques of equation solving learned here in order to answer questions about these phenomena.
Points to Consider
 What kinds of situations might be modeled with exponential functions or logarithmic functions?
 What restrictions are there on the domain and range of data if we use these functions as models?
 When might an exponential or logarithmic equation have no solution?
 What are the advantages and disadvantages of using a graphing calculator to solve exponential and logarithmic equations?
Review Questions
For questions 1  5, solve each equation using algebraic methods. Give an exact solution.
 2 (5^{x} ^{ 4}) + 7 = 43
 4^{x} = 7^{3}^{x} ^{ 5}
 log(5x + 200) + log 2 = 3
 log_{3} (4x + 5)  log_{3} x = 2
 ln (4x + 1)  ln (2x) = 3
 Use a graphing utility to solve the equation in #4.
 Use a graphing utility to solve the equation log(x^{2} 3) = log (x + 5)
 In example 3b, the solution to the log equation log_{2}(3x + 8) + 1 = log_{3} (10  x) was found to be x ≈  1.87. One student read this example, and wondered how the value of x could be negative, given that you cannot take a log of a negative number. How would you explain to this student why the solution is valid?
 The data set below represents a hypothetical situation: You invest $2000 in a money market account, and you do not invest more money or withdraw any from the account. Table 3

Time since you invested (in years) Amount in account 0 2,000 2 2200 5 2500 10 3300 20 4500  a. Use a graphing utility to find an exponential model for the data. b. Use your model to estimate the value of the account after the 8
 ^{th}
 year. c. At this rate, much money would be in the account after 30 years? d. Explain how your estimate in part c might be inaccurate. (What might happen after 20 years?)
 The data set below represents the growth of a plant. Table 4

Time since planting (days) Height of plant (inches) 1 .2 4 .5 5 .57 10 1.2 12 1.3 14 1.4  a. Use a graphing utility to find a logarithmic equation to model the data. b. Use your model to estimate the height of the plant after 15 days. Compare this estimate to the trend in the data. c. Give an example of an
 x
 value for which the model does not make sense.
 In the lesson, the equation log(5  x) + 1 = log x was solved using a graph. Solve this equation algebraically in order to (a) verify the approximate solution found in the lesson and (b) give an exact solution.
Review Answers
 \begin{align*}log_5 18 + 4\end{align*} or \begin{align*}\frac{log 18} {log 5} + 4\end{align*}
 \begin{align*}\frac{5 log 7} {log 4  3 log 7}\end{align*}
 x = 60
 x = 1
 \begin{align*}x = \frac{1} {2e^3  4}\end{align*}
 The function y = log_{3} (4x + 5)  log_{3} x intersects the line y = 2 at the point (1, 2)
 The graphs intersect twice, giving 2 solutions: x ≈ 2.37, x ≈ 3.37
 The value of can be negative as long as the argument of the log is positive. In this equation, the arguments are 3x+8 and 10x. Neither expression takes on a negative value for x ≈ 1.87
 a. y = 2045.405(1.042)^{x} b. About $2840 c. About $7003 d. After that much time, you may decide to withdraw the money to spend or to invest in something with more potential for growth.
 a. y = 0.0313 + .4780 ln x b. The model gives 1.32 inches. The data would suggest the plant is at least 1.4 inches tall. c. The model does not make sense for negative x values. Also, at some point the plant could die. This reality puts an upper bound on x.
 log (5  x) + 1 = log x log (5  x)  log x + 1 = 0 log(5  x)  log x = 1 \begin{align*}log \left ( \frac{5x}{x} \right )=1 \end{align*} \begin{align*}10^{1} = \frac{5  x} {x}\end{align*} \begin{align*}0.1 = \frac{5  x} {x}\end{align*} 0.1 x = 5  x 1.1 x = 5 \begin{align*}x = \frac{5} {1.1} = \frac{50} {11} = 4 \frac{6} {11} = 5.\overline{54}\end{align*}
Vocabulary
 Extraneous solution
 An extraneous solution is a solution to an equation used to solve an initial equation that is not a solution to the initial equation. Extraneous solutions occur when solving certain kinds of equations, such as log equations, or square root equations.
 Extrapolation
 To extrapolate from data is to create new data points, or to predict, outside of the domain of the data set.
 Interpolation
 To interpolate is to create new data points, or to predict, within the domain of the data set, but for points not in the original data set.