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Section 15.1 Solving Log and Exponential Equations

You already know a lot about solving equations. In this section, we will see how to use our new tool of logarithms to solve equations that involve a variable in the exponent, or inside of a log itself.

Example 15.1. Motivational Example.

Let's say we want to solve the equation

\begin{equation*} x+3=7 \end{equation*}

Well, we need to get \(x\) by itself. The \(3\) is in the way of that, so we need to move it. Since the \(3\) is attached to the \(x\) by addition, we will use subtraction to move it, since subtraction cancels addition.

\begin{align*} x+3\amp= 7\\ x+3 {\color{red}{-3}}\amp=7{\color{red}{-3}}\\ x \amp=4 \end{align*}

Let's do another. Suppose we want to solve the equation

\begin{equation*} 3x=7 \end{equation*}

Well, we need to get \(x\) by itself. The \(3\) is in the way of that, so we need to move it. Since the \(3\) is attached to the \(x\) by multiplication, we will use division to move it, since division cancels multiplication.

\begin{align*} 3x\amp= 7\\ \frac{3x}{\color{red}{3}}\amp=\frac{7}{\color{red}{3}}\\ x \amp=\frac{7}{3} \end{align*}

One more. Suppose we want to solve the equation

\begin{equation*} x^3=7 \end{equation*}

Well, we need to get \(x\) by itself. The \(3\) is in the way of that, so we need to move it. Since the \(3\) is attached to the \(x\) as the power, we will use a root to move it, since the cube root cancels a cube.

\begin{align*} x^3\amp= 7\\ \sqrt[{\color{red}{3}}]{x^3}\amp=\sqrt[{\color{red}{3}}]{7}\\ x \amp=\sqrt[3]{7} \end{align*}

In each example above, we had to ask how the number was attached to \(x\text{,}\) and then use the operation that cancels that. Now, we want to be able to solve equations like

\begin{equation*} 3^x = 7 \end{equation*}

This is an exponential function. As we learned in Section 13.1, logs and exponentials cancel, as long as the bases match. So, to solve this equation, we will take the log base 3 of both sides.

\begin{align*} 3^x\amp= 7\\ {\color{red}{\log_3}}\left(3^x\right)\amp={\color{red}{\log_3}}\left(7\right)\\ x \amp=\log_3(7) \end{align*}

The only other thing we have to be careful of is our order of operations. Don't forget that we are solving equations, we do the order of operations backwards. So, if we're solving an equation, we work in the following order:

  1. Addition/subtraction

  2. Multiplication/division

  3. Exponents/roots

  4. Groups/parentheses

Since logs and exponentials are functions, they are a type of group.

Example 15.2.

Suppose we want to solve the following equation

\begin{equation*} 2\cdot 5^{x-4}-8=6 \end{equation*}

The variable is in an exponent, so we will need a log at some point, but we have some other operations to deal with first. According to our order of operations (which we do backwards when we are solving an equation), we have to deal with the addition/subtraction first. The 4 is also buried in an exponent, so it's actually part of a group, so we can't get to that one yet. But we can do some other operations outside of the exponential.

\begin{align*} 2\cdot 5^{x-4}-8\amp=6\\ 2\cdot 5^{x-4}-8{\color{red}{+8}}\amp=6{\color{red}{+8}}\\ 2\cdot 5^{x-4}\amp=14\\ \frac{2\cdot 5^{x-4}}{\color{red}{2}}\amp=\frac{14}{\color{red}{2}}\\ 5^{x-4}\amp=7 \end{align*}

Now we are ready to use the log to cancel our exponential. Since the exponential is base 5, we are going to use the log with the same base.

\begin{align*} {\color{red}{\log_5}}\left(5^{x-4}\right)\amp={\color{red}{\log_5}}\left(7\right)\\ x-4 \amp= \log_5(7) \end{align*}

Finally, now that the 4 is out of the exponent, we can subtract to get our final answer:

\begin{align*} x-4{\color{red}{+4}} \amp= \log_5(7){\color{red}{+4}} \\ x\amp= \log_5(7)+4 \end{align*}

Checkpoint 15.3.

Solve the following exponential equation:

\begin{equation*} e^{3x-1}+5=9 \end{equation*}
Answer.
\(x = \frac{\ln(4)+1}{3}\)
Solution.

The variable is in an exponent, so we will need a log at some point, but we have some other operations to deal with first. According to our order of operations (which we do backwards when we are solving an equation), we have to deal with the addition/subtraction first. The 3 and the 1 are also buried in an exponent, so they are actually part of a group, so we can't get to those yet. But we can do some other operations outside of the exponential.

\begin{align*} e^{3x-1}+5\amp=9\\ e^{3x-1}+5{\color{red}{-5}}\amp=9{\color{red}{-5}}\\ e^{3x-1}\amp=4 \end{align*}

Now we are ready to use the log to cancel our exponential. Since the exponential is base \(e\text{,}\) we are going to use the log with the same base. Remember that log base \(e\) is just the natural logarithm.

\begin{align*} {\color{red}{\ln}}\left(e^{3x-1}\right)\amp={\color{red}{\ln}}\left(4\right)\\ 3x-1 \amp= \ln(4) \end{align*}

Finally, now that everything is out of the exponent, we can do the last few operations to get our final answer:

\begin{align*} 3x-1{\color{red}{+1}} \amp= \ln(4){\color{red}{+1}} \\ 3x\amp= \ln(4)+1\\ \frac{3x}{\color{red}{3}}\amp= \frac{\ln(4)+1}{\color{red}{3}}\\ x \amp = \frac{\ln(4)+1}{3} \end{align*}

Example 15.4.

We can use the same idea to solve an equation that has a log in it to begin with. For example, suppose we want to solve the following equation:

\begin{equation*} 4\log_5(x+7)-3=9 \end{equation*}

The variable is in a log, so we will need an exponential at some point, but we have some other operations to deal with first. According to our order of operations (which we do backwards when we are solving an equation), we have to deal with the addition/subtraction first. The 7 is buried inside the log, so it is actually part of a group, so we can't get to that yet. But we can do some other operations outside of the log.

\begin{align*} 4\log_5(x+7)-3\amp=9\\ 4\log_5(x+7)-3{\color{red}{+3}}\amp=9{\color{red}{+3}}\\ 4\log_5(x+7)\amp=12\\ \frac{4\log_5(x+7)}{\color{red}{4}}\amp=\frac{12}{\color{red}{4}}\\ \log_5(x+7)\amp=3 \end{align*}

Now we are ready to use the exponential to cancel our log. Since the log is base \(5\text{,}\) we are going to use the exponential with the same base. In other words, each side of our equation will become the exponent on 5.

\begin{align*} {\color{red}{5}}^{\log_5(x+7)}\amp={\color{red}{5}}^3\\ x+7\amp=125 \end{align*}

Finally, now that everything is out of the log, we can do the last operation to get our final answer:

\begin{align*} x+7{\color{red}{-7}} \amp= 125{\color{red}{-7}} \\ x\amp= 118 \end{align*}

Checkpoint 15.5.

Solve the following equation:

\begin{equation*} 3\log_7(8x-5)+4=10 \end{equation*}
Answer.

\(x=\frac{27}{4}\)

Solution.

The variable is in a log, so we will need an exponential at some point, but we have some other operations to deal with first. According to our order of operations (which we do backwards when we are solving an equation), we have to deal with the addition/subtraction first. The 8 and 5 are buried inside the log, so they are actually part of a group, so we can't get to those yet. But we can do some other operations outside of the log.

\begin{align*} 3\log_7(8x-5)+4\amp=10\\ 3\log_7(8x-5)+4{\color{red}{-4}}\amp=10{\color{red}{-4}}\\ 3\log_7(8x-5)\amp=6\\ \frac{3\log_7(8x-5)}{\color{red}{3}}\amp=\frac{6}{\color{red}{3}}\\ \log_7(8x-5)\amp=2 \end{align*}

Now we are ready to use the exponential to cancel our log. Since the log is base \(7\text{,}\) we are going to use the exponential with the same base. In other words, each side of our equation will become the exponent on 7.

\begin{align*} {\color{red}{7}}^{\log_7(8x-5)}\amp={\color{red}{7}}^2\\ 8x-5\amp=49 \end{align*}

Finally, now that everything is out of the log, we can do the last operation to get our final answer:

\begin{align*} 8x-5{\color{red}{+5}} \amp= 49{\color{red}{+5}} \\ 8x\amp= 54\\ \frac{8x}{\color{red}{8}}\amp= \frac{54}{\color{red}{8}}\\ x\amp=\frac{54}{8}=\frac{27}{4} \end{align*}