Answer
$x=21$
Work Step by Step
RECALL:
$\log_b{x}=y \longrightarrow b^y=x.$
Use the rule above to write the given equation in exponential form, then solve the equation:
$\begin{array}{ccc}
&5^2 &= &x+4
\\&25 &= &x+4
\\&25-4 &= &x+4-4
\\&21 &= &x
\end{array}.$
Thus, $x=21$.
