Answer
$x=\dfrac{\pm \sqrt{2e^{\frac{1}{2}}}}{2}$
Work Step by Step
Divide $2$ to both sides:
$$\ln{\left(2x^2\right)}=\frac{1}{2}$$
Recall:
$$\ln{a}=y \longleftrightarrow e^y=a$$
Use the definition above to obtain:
\begin{align*}
\ln{\left(2x^2\right)}&=\frac{1}{2}\\\\
e^{\frac{1}{2}}&=2x^2\\\\
\frac{e^{\frac{1}{2}}}{2}&=\frac{2x^2}{2}\\\\
\frac{e^{\frac{1}{2}}}{2}&=x^2\\\\
\pm\sqrt{\frac{e^{\frac{1}{2}}}{2}}&=x\\\\
\pm\sqrt{\frac{e^{\frac{1}{2}}}{2}\cdot \frac{2}{2}}&=x\\\\
\pm\sqrt{\frac{2e^{\frac{1}{2}}}{4}}&=x\\\\
\dfrac{\pm\sqrt{2e^{\frac{1}{2}}}}{2}&=x\\\\
\end{align*}
Check:
\begin{align*}
2\ln{\left(2\left(\frac{\sqrt{2e^{\frac{1}{2}}}}{2}\right)^2\right)}&\stackrel{?}=1\\\\
2\ln{\left(2\left(\frac{2e^{\frac{1}{2}}}{4}\right)\right)}&\stackrel{?}=1\\\\
2\ln{\left(e^{\frac{1}{2}}\right)}&\stackrel{?}=1\\\\
2\left(\frac{1}{2}\right)&\stackrel{?}=1\\\\
1&\stackrel{\checkmark}=1
\end{align*}
\begin{align*}
2\ln{\left(2\left(\frac{-\sqrt{2e^{\frac{1}{2}}}}{2}\right)^2\right)}&\stackrel{?}=1\\\\
2\ln{\left(2\left(\frac{2e^{\frac{1}{2}}}{4}\right)\right)}&\stackrel{?}=1\\\\
2\ln{\left(e^{\frac{1}{2}}\right)}&\stackrel{?}=1\\\\
2\left(\frac{1}{2}\right)&\stackrel{?}=1\\\\
1&\stackrel{\checkmark}=1
\end{align*}
