Answer
$x=2$
Work Step by Step
We know that if $\log_a {y}=x$ then $a^x=y$.
Hence, if $\log_x {(4)}=2$, then
\begin{align*}x^2=4\\
\sqrt{x^2}=\pm\sqrt{4}\\
x=\pm2\end{align*}
But by the definition of the logarithmic function the base of the logarithm must be positive, hence $x=2.$
