Answer
$x=5$
Work Step by Step
First, the solutions must satisfy $\left\{\begin{array}{l}
x-3\gt 0\\
x+3\gt 0
\end{array}\right.\quad \Rightarrow x\gt 3\qquad (*)$
in order for the equation to be defined.
LHS: Apply$ \quad\log_{a}(MN)=\log_{a}M+\log_{a}N$
RHS: Apply $\quad \log_{2}2^{4}=4$
$\log_{2}[ (x-3)(x+3)]=\log_{2}16$
... apply the principle of logarithmic equality
$(x-3)(x+3)=16$
$x^{2}-9=16$
$x^{2}=25$
Possible solutions:
$ x=-5\qquad$... does not satisfy (*), not a solution.
$x=5\qquad $... satisfies (*), and is a valid solution.
