# 3.6: Exponential Equations

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When you were first learning equations, you learned the rule that whatever you do to one side of an equation, you must also do to the other side so that the equation stays in balance. The basic techniques of adding, subtracting, multiplying and dividing both sides of an equation worked to solve almost all equations up until now. With logarithms, you have more tools to isolate a variable. Consider the following equation and ask yourself: why is \(x=3\) ? Logically it makes sense that if the bases match, then the exponents must match as well, but how can it be shown for examples like this one?

\(1.79898^{2 x}=1.79898^{6}\)

## Solving Exponential Equations

A common technique for solving equations with unknown variables in exponents is to take the log of the desired base of both sides of the equation. Then, you can use properties of logs to simplify and solve the equation.

Take the following equation. To solve for \(t\), you should first simplify the expression as much as possible and then take the natural log of both sides.

\(9,000=300 \cdot \frac{(1.06)^{t}-1}{0.06}\)

\(\begin{aligned} 30 &=\frac{(1.06)^{t}-1}{0.06} \\ 1.8 &=(1.06)^{t}-1 \\ 2.8 &=1.06^{t} \\ \ln 2.8 &=\ln \left(1.06^{t}\right)=t \cdot \ln (1.06) \\ t &=\frac{\ln 2.8}{\ln 1.06} \approx 17.67 \text { years} \end{aligned}\)

It does not matter what base you use in this situation as long as you use the same base on both sides. Choosing natural log allows you to use a calculator to finish the problem.

Note that this type of equation is common in financial mathematics. The equation above represents the unknown amount of time it will take you to save $9,000 in a savings account if you save $300 at the end of each year in an account that earns 6% annual compound interest.

The other good base to use is base \(10 .\) When solving the following equation for \(x: 16^{x}=25\), you will need to use a calculator to get the final answer and your calculator can handle base 10 as well. First take the log of both sides. Then, use log properties and your calculator to help.

\(\begin{aligned} 16^{x} &=25 \\ \log 16^{x} &=\log 25 \\ x \log 16 &=\log 25 \\ x &=\frac{\log 25}{\log 16} \\ x &=1.16 \end{aligned}\)

## Examples

Earlier, you were asked how to show that if the bases match in an equation, the exponents should match. In the equation, logs can be used to reduce the equation to \(2 x=6\).

\(1.79898^{2 x}=1.79898^{6}\)

Take the log of both sides and use the property of exponentiation of logs to bring the exponent out front.

\(\begin{aligned} \log 1.79898^{2 x} &=\log 1.79898^{6} \\ 2 x \cdot \log 1.79898 &=6 \cdot \log 1.79898 \\ 2 x &=6 \\ x &=3 \end{aligned}\)

Solve the following equation for all possible values of \(x:(x+1)^{x-4}-1=0\)

\((x+1)^{x-4}-1=0\)

\((x+1)^{x-4}=1\)

Case 1 is that \(x+1\) is positive in which case you can take the log of both sides.

\(\begin{aligned} \log (x+1)^{(x-4)} &=\log 1 \\(x-4) \cdot \log (x+1) &=0 \\ x-4=0 & \text { or } \log (x+1)=0 \\ x=& 4 \text { or }(x+1)=10^{0}=1 \\ & x=4,0 \end{aligned}\)

Note that \(\log 1=0\)

Case 2 is that \((x+1)\) is negative 1 and raised to an even power. This happens when \(x=-2\).

\(\begin{aligned}(x+1)^{(x-4)} &=1 \\(-2+1)^{(-2-4)}-1 &=(-1)^{-6}-1 \\ &=\frac{1}{(-1)^{6}}-1 \\ &=0 \end{aligned}\)

The reason why this exercise is included is because you should not fall into the habit of assuming that you can take the log of both sides of an equation. It is only valid when the argument is strictly positive. For example, \(\log (-2+1)^{(-2-4)}=\log (-1)\) is not possible.

Light intensity, measured in lumens, can be described by the relationship between \(i\) for intensity and \(d\) for depth in feet as it travels at specific depths of water in a swimming pool. What is the intensity of light at 10 feet?

\(\log \left(\frac{i}{12}\right)=-0.0145 \cdot d\)

Given \(d=10\), solve for \(i\) measured in lumens.

\(\begin{aligned} \log \left(\frac{i}{12}\right) &=-0.0145 \cdot d \\ \log \left(\frac{i}{12}\right) &=-0.0145 \cdot 10 \\ \log \left(\frac{i}{12}\right) &=-0.145 \\\left(\frac{i}{12}\right) &=10^{-0.145} \\ i &=12 \cdot 10^{-0.145} \approx 8.594 \end{aligned}\)

Solve the following equation for all possible values of \(x\).

\(\frac{e^{x}-e^{-x}}{3}=14\)

First solve for \(e^{x}\),

\(\begin{aligned} \frac{e^{x}-e^{-x}}{3} &=14 \\ e^{x}\left(e^{x}-e^{-x}\right) &=(42) e^{x} \\ e^{2 x}-1 &=42 e^{x} \\\left(e^{x}\right)^{2}-42 e^{x}-1 &=0 \end{aligned}\)

Let \(u=e^{x}\)

\(u^{2}-42 u-1=0\)

\(u=\frac{-(-42) \pm \sqrt{(-42)^{2}-4 \cdot 1 \cdot(-1)}}{2 \cdot 1}=\frac{42 \pm \sqrt{1768}}{2} \approx 42.023796,-0.0237960\)

Note that the negative result is extraneous because \(e^{x}\) must be greater than zero, so you only proceed in solving for \(x\) for one result.

\(\begin{aligned} e^{x} & \approx 42.023796 \\ x & \approx \ln 42.023796 \approx 3.738 \end{aligned}\)

Solve the following equation for all possible values of \(x:\left(\log _{2} x\right)^{2}-\log _{2} x^{7}=-12\).

In calculus it is common to use a small substitution to simplify the problem and then substitute back later. In this case let \(u=\log _{2} x\). Notice that this is a quadratic problem.

\(\begin{aligned}\left(\log _{2} x\right)^{2}-7 \log _{2} x+12 &=0 \\ u^{2}-7 u+12 &=0 \\(u-3)(u-4) &=0 \\ u &=3,4 \end{aligned}\)

Now substitute back and solve for \(x\) in each case.

\(\log _{2} x=3 \leftrightarrow 2^{3}=x=8\)

\(\log _{2} x=4 \leftrightarrow 2^{4}=x=16\)

Review

Solve each equation for \(x\). Round each answer to three decimal places.

1. \(4^{x}=6\)

2. \(5^{x}=2\)

3. \(12^{4 x}=1020\)

4. \(7^{3 x}=2400\)

5. \(2^{x+1}-5=22\)

6. \(5 x+12^{x}=5 x+7\)

7. \(2^{x+1}=2^{2 x+3}\)

8. \(3^{x+3}=9^{x+1}\)

9. \(2^{x+4}=5^{x}\)

10. \(13 \cdot 8^{0.2 x}=546\)

11. \(b^{x}=c+a\)

12. \(32^{x}=0.94-.12\)

Solve each log equation by using log properties and rewriting as an exponential equation.

13. \(\log _{3} x+\log _{3} 5=2\)

14. \(2 \log x=\log 8+\log 5-\log 10\)

15. \(\log _{9} x=\frac{3}{2}\)

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