Answer
$x=\sqrt{e^4+1}\approx 7.456$
Work Step by Step
Use the Product Rule to obtain:
\begin{align*}
\ln{\left((x+1)(x-1)\right)}&=4\\
\ln{\left(x^2-1\right)}&=4
\end{align*}
Recall:
$$\ln{x}=n \longrightarrow e^n=x$$
Use the definition above to obtain:
\begin{align*}
\ln{\left(x^2-1\right)}&=4\\\\
e^4&=x^2-1\\\\
e^4+1&=x^2\\\\
\pm \sqrt{e^4+1}&=\sqrt{x^2}\\\\
x&=\pm\sqrt{e^4+1}\\\\
x&\approx\pm7.456
\end{align*}
However, since in $\ln{x}$, $x$ must be greater than zero, then $x=-7.456$ is an extraneous solution.
Therefore, the solution is:
$x=\sqrt{e^4+1}\approx 7.456$
