Exponential and logarithmic equations

Example 1: Solve ${3^{2x-1}=27}$.

Solution:

Write ${27}$ as a power of ${3}$: ${27=3^3}$. Then set the exponents equal:

$\begin{align*}3^{2x-1}&=3^3\\2x-1&=3\\x&=2\end{align*}$

Check: ${3^{2(2)-1}=3^3=27}$.

Example 2: Solve ${2^x=7}$.

Solution:

Take ${\ln}$ of both sides, and use the power law:

$\begin{align*}\ln 2^x&=\ln 7\\x\ln 2&=\ln 7\\x&=\dfrac{\ln 7}{\ln 2}\approx 2.807\end{align*}$

This makes sense: ${7}$ is between ${2^2=4}$ and ${2^3=8}$, so ${x}$ is between ${2}$ and ${3}$.

Example 3: Solve ${5\cdot 3^x=40}$.

Solution:

First isolate the power: divide both sides by ${5}$. Then take ${\ln}$:

$\begin{align*}3^x&=8\\x\ln 3&=\ln 8\\x&=\dfrac{\ln 8}{\ln 3}\approx 1.893\end{align*}$

Example 4: Solve ${e^{3-2x}=4}$.

Solution:

Take ${\ln}$ of both sides. The left side becomes ${3-2x}$:

$\begin{align*}3-2x&=\ln 4\\-2x&=\ln 4-3\\x&=\dfrac{3-\ln 4}{2}\approx 0.807\end{align*}$

Example 5: Solve ${e^{2x}-e^x-6=0}$.

Solution:

Since ${e^{2x}=\left(e^x\right)^2}$, this is a quadratic equation in ${e^x}$ (see other types of equations). Factor it:

${\left(e^x-3\right)\left(e^x+2\right)=0}$

So ${e^x=3}$ or ${e^x=-2}$.

The first gives ${x=\ln 3\approx 1.099}$.

The second has no solution, because ${e^x}$ is always positive.

The only solution is ${x=\ln 3}$.

Example 6: Solve ${2^{x+1}=3^x}$.

Solution:

The bases are different. Take ${\ln}$ of both sides, and use the power law on both sides:

${(x+1)\ln 2=x\ln 3}$

Multiply out. Then collect the ${x}$-terms on one side, and factor out ${x}$:

$\begin{align*}x\ln 2+\ln 2&=x\ln 3\\\ln 2&=x\ln 3-x\ln 2\\\ln 2&=x(\ln 3-\ln 2)\end{align*}$

Divide:

$x=\dfrac{\ln 2}{\ln 3-\ln 2}\approx 1.710$

Check: ${2^{2.710}\approx 6.54}$ and ${3^{1.710}\approx 6.54}$.

Example 7: Solve ${\log_2(x+3)=4}$.

Solution:

Write it in exponential form:

$\begin{align*}x+3&=2^4=16\\x&=13\end{align*}$

Check: ${\log_2(13+3)=\log_2 16=4}$.

Example 8: Solve ${4+3\log(2x)=16}$.

Solution:

First isolate the logarithm:

$\begin{align*}3\log(2x)&=12\\\log(2x)&=4\end{align*}$

The base is ${10}$. Write in exponential form:

$\begin{align*}2x&=10^4=10{,}000\\x&=5000\end{align*}$

Check: ${\log 10{,}000=4}$, and ${4+3(4)=16}$.

Example 9: Solve ${\log(x+2)+\log(x-1)=1}$.

Solution:

Combine the logarithms with the product law. Then write in exponential form:

$\begin{align*}\log\big((x+2)(x-1)\big)&=1\\(x+2)(x-1)&=10\end{align*}$

Multiply out, and solve the quadratic:

$\begin{align*}x^2+x-2&=10\\x^2+x-12&=0\\(x+4)(x-3)&=0\end{align*}$

So ${x=-4}$ or ${x=3}$.

Check ${x=-4}$: ${\log(-4+2)=\log(-2)}$ is undefined. So ${x=-4}$ is extraneous.

Check ${x=3}$: ${\log 5+\log 2=\log 10=1}$. It checks.

The only solution is ${x=3}$.

Example 10: You invest ${\$5000}$ at ${5\%}$ interest per year, compounded monthly. How long does it take for the money to double?

Solution:

Use the compound interest formula from exponential functions. The money doubles when ${A=10{,}000}$:

$5000\left(1+\dfrac{0.05}{12}\right)^{12t}=10{,}000$

Divide both sides by ${5000}$, and take ${\ln}$:

$\begin{align*}\left(1+\dfrac{0.05}{12}\right)^{12t}&=2\\12t\ln\left(1+\dfrac{0.05}{12}\right)&=\ln 2\end{align*}$

Divide both sides by ${12\ln\left(1+\dfrac{0.05}{12}\right)}$:

$t=\dfrac{\ln 2}{12\ln\left(1+\frac{0.05}{12}\right)}\approx 13.89$

The money doubles in about ${13.9}$ years. The starting amount does not matter: ${\$100}$ would also double in ${13.9}$ years.