Section 7.6 Exponential Equations
Logarithms are a useful tool for expressing solutions to exponential equations. We’ll consider some examples in this section.
Example 7.31 . Solving an Exponential Equation.
Consider the exponential equation
\begin{equation*}
5^x = 10\text{.}
\end{equation*}
First observe that this equation must have a unique solution. This is because the exponential function \(f(x)=5^x\) is one-to-one. In the figure below it appears that the graph of \(y =10\) crosses \(y=5^x\) exactly once somewhere near \(x\approx 1.5\text{.}\)
Figure 7.32. Plot of \(y = 10\) and \(y = 5^x\text{.}\)
On the one hand, we can rewrite the equation in logarithmic form to obtain
\begin{equation*}
5^{\boxed{x}} = 10 \quad \Rightarrow \quad \boxed{x} = \log_5(10)\text{.}
\end{equation*}
So that the solution is a logarithm. Unfortunately, your calculator is unlikely to have the ability to estimate base 5 logarithms.
Alternatively, we can express the solution in terms of the natural logarithm. Apply the natural logarithm to both sides to obtain an equivalent equation:
\begin{gather*}
5^x = 10\\
\ln(5^x) = \ln(10)
\end{gather*}
Use the power property of logarithms and solve for \(x\)
\begin{gather*}
x \cdot \ln(5) = \ln(10)\\
x = \frac{\ln(10)}{\ln(5)}
\end{gather*}
Thus, it must be that our base 5 answer is equal to this ratio of natural logarithms:
\begin{equation*}
x = \log_5(10) = \frac{\ln(10)}{\ln(5)}
\end{equation*}
We can now estimate with a calculator by diving the two logs.
\begin{equation*}
x = \log_5(10) \approx 1.431.
\end{equation*}
In exponential form,
\begin{equation*}
5^{1.431} \approx 10.
\end{equation*}
The previous example illustrates the following change of base formula. In principle, one only needs the natural logarithm to solve exponential equations.
Theorem 7.33 . Change of Base for Logarithms.
For \(a\gt 0\text{,}\)
\begin{equation*}
\log_a(x) = \frac{\ln(x)}{\ln(a)}.
\end{equation*}
Example 7.34 . Solving an Exponential Equation.
Solve
\begin{equation*}
\frac{1}{5^x} - 10 = 2.
\end{equation*}
Solution .
First, isolate the exponential term
\begin{gather*}
\frac{1}{5^x} - 10 = 2\\
\frac{1}{5^x} = 12\\
5^{-x} = 12
\end{gather*}
Now apply the natural logarithm to both sides and use the power property of logarithms
\begin{gather*}
\ln\left(5^{-x}\right) = \ln(12)\\
-x \cdot \ln\left(5\right) = \ln(12)\\
-x = \frac{\ln(12)}{\ln(5)}\\
x = \frac{\ln(12)}{\ln(5)}
\end{gather*}
which can now be estimated with a calculator
\begin{equation*}
x \approx 1.544
\end{equation*}
Example 7.35 . Solving an Exponential Equation.
The value in dollars of a car after \(t\) years since purchase depreciates according to the formula
\begin{equation*}
A(t) = 19000\times (0.65)^t + 1000.
\end{equation*}
How log will it take before the car is worth a quarter of its original purchase price?
Solution .
The purchase price is
\begin{equation*}
A(0)=19000\times (0.65)^0 + 1000 = 19000+1000 = 20000.
\end{equation*}
A quarter of this amount is \(20000/4 = 5000\) dollars. We want to solve the equation
\begin{equation*}
A(t) = 5000.
\end{equation*}
So,
\begin{gather*}
19000\times (0.65)^t + 1000 = 5000\\
19000\times (0.65)^t = 4000\\
(0.65)^t = \frac{4000}{19000}\\
(0.65)^t = \frac{4}{19}\\
\ln\left((0.65)^t\right) = \ln\left(\frac{4}{19}\right)\\
t\cdot \ln\left(0.65\right) = \ln\left(\frac{4}{19}\right)\\
t = \frac{\ln\left(\frac{4}{19}\right)}{\ln(0.65)}
\end{gather*}
Estimating with a calculator,
\begin{equation*}
t \approx 3.62.
\end{equation*}
We expect it to take 3.63 years for the car to depreciate to a quarter of its original purchase price.
