Answer
$x=\displaystyle \frac{17}{5}$
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.
On the RHS, apply$ \quad \log_{2}2^{4}=4$
$\log_{2}(x+3)=\log_{2}16+\log_{2}(x-3)$
... apply $\qquad \log_{a}(MN)=\log_{a}M+\log_{a}N$
$\log_{2}(x+3)=\log_{2}[16(x-3)]$
... apply the principle of logarithmic equality
$x+3=16(x-3)$
$ x+3=16x-48\qquad$... add $48-x$
$51=15x$
$x=\displaystyle \frac{51}{15}=\frac{17}{5}\quad $... satisfies (*), and is a valid solution.
