Answer
$x=0$
Work Step by Step
Multiply $e^x$ to both sides to obtain:
\begin{align*}
e^x\left(\frac{e^x+e^{-x}}{2}\right)&=1(e^x)\\\\
\frac{e^x\cdot e^x+e^x\cdot e^{-x}}{2}&=e^x\\\\
\frac{e^{2x}+e^{x+(-x)}}{2}&=e^x\\\\
\frac{e^{2x}+e^0}{2}&=e^x\\\\
\frac{e^{2x}+1}{2}&=e^x\\\\
\end{align*}
Multiply $2$ to both sides to obtain:
\begin{align*}
2\left(\frac{e^{2x}+1}{2}\right)&=e^x(2)\\\\
e^{2x}+1&=2e^x\\\\
e^{2x}+1-2e^x&=0\\\\
e^{2x}-2e^x+1&=0\\\\
\end{align*}
Factor the trinomial to obtain:
\begin{align*}
(e^x-1)(e^x-1)&=0\\
(e^x-1)^2&=0\end{align*}
Use the Zero-Product Prperty by equating each unique factor to zero, then solve the equation to obtain:
\begin{align*}
e^x-1&=0\\
e^x&=1\\
\ln{(e^x)}&=\ln1\\
x&=0
\end{align*}
Therefore, the solution is $x=0$.
