Answer
\begin{align*}
x&=\ln{(3-2\sqrt2)}\approx-1.763\\
x&=\ln{(3+2\sqrt2)}\approx1.763
\end{align*}
Work Step by Step
Multiply $e^x$ to both sides to obtain:
\begin{align*}
e^x\left(\frac{e^x+e^{-x}}{2}\right)&=3(e^x)\\\\
\frac{e^x\cdot e^x+e^x\cdot e^{-x}}{2}&=3e^x\\\\
\frac{e^{2x}+e^{x+(-x)}}{2}&=3e^x\\\\
\frac{e^{2x}+e^0}{2}&=3e^x\\\\
\frac{e^{2x}+1}{2}&=3e^x\\\\
\end{align*}
Multiply $2$ to both sides to obtain:
\begin{align*}
2\left(\frac{e^{2x}+1}{2}\right)&=(3e^x)(2)\\\\
e^{2x}+1&=6e^x\\\\
e^{2x}+1-6e^x&=0\\\\
e^{2x}-6e^x+1&=0\\\\
e^{2x}-6e^x&=-1\\\\
\end{align*}
Complete the square by adding $\left(\frac{-6}{2}\right)^2=(-3)^2=9$ to both sides to obtain:
\begin{align*}
e^{2x}-6x+9&=-1+9\\
e^{2x}-6x+9&=8\\
(e^x-3)^2&=8
\end{align*}
Take the square root of both sides to obtain:
\begin{align*}
\sqrt{(e^x-3)^2}&=\pm\sqrt8\\
e^x-3&=\pm\sqrt{4(2)}\\
e^x-3&=\pm2\sqrt2\\
e^x&=3\pm2\sqrt2
\end{align*}
Take the natural log of both sides to obtain:
\begin{align*}
\ln{e^x}&=\ln{3\pm2\sqrt2}\\
x&=\ln{(3\pm2\sqrt2)}\\
\end{align*}
Splitting the answers gives:
\begin{align*}
x&=\ln{(3-2\sqrt2)}\approx-1.763\\
x&=\ln{(3+2\sqrt2)}\approx1.763
\end{align*}
