Answer
$x=\dfrac{e^2}{8}$
Work Step by Step
Recall:
(1) $n\cdot \ln{a}=\ln{a^n}$
(2) $\ln{a}+\ln{b} = \ln{(ab)}$
Use rule (1) above to obtain:
\begin{align*}
\ln{x^2}+\ln{4^3}&=4\\
\ln{x^2}+\ln{64}&=4\\
\end{align*}
Use rule (2) above to obtain:
\begin{align*}
\ln{\left(x^2\cdot 64\right)}&=4\\
\ln{\left(64x^2\right)}&=4\\
\end{align*}
Recall:
$\ln{x}=a \longleftrightarrow e^a=x$
Hence,
\begin{align*}
\ln{(64x^2)}=4 &\longrightarrow 64x^2=e^4\\\\
x^2=\frac{e^4}{64}\\\\
x&=\pm\sqrt{\frac{e^4}{64}}\\\\
x&=\pm\frac{e^2}{8}
\end{align*}
However, since $x$ in $\ln{x}$ must be non-negative, $x \ne -\frac{e^2}{8}$.
Thus, the solution is $x=\dfrac{e^2}{8}$.
Check:
\begin{align*}
2\ln{\left(\frac{e^2}{8}\right)} + 3\ln{4}&=4\\\\
\ln{\left(\left[\frac{e^2}{8}\right]^2\right)}+\ln{4^3}&=4\\\\
\ln{\left(\frac{e^4}{64}\right)}+\ln{64}&=4\\\\
\ln{\left(\frac{e^4}{64} \cdot 64\right)}&=4\\\\
\ln{e^4}&=4\\\\
4&=4
\end{align*}
