## Precalculus (6th Edition) Blitzer

$x=21$
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, $x=21$.