Answer
$x=2$
Work Step by Step
RECALL:
$\log_a{x} = y \longrightarrow a^y=x, a\gt0, a\ne 1$
Use the definition above to obtain:
$\log_x{4}=2
\\
\longrightarrow x^2=4$
Take the square root of both sides to obtain:
$x = \pm \sqrt{4}
\\x = \pm \sqrt{2^2}
\\x = \pm 2$
Since the base of a logarithm must be positive, $x$ cannot be $-2$.
Thus, the only solution to the given equation is $x=2$.
