Answer
$x=\displaystyle \frac{83}{15}$
Work Step by Step
First, the solutions must satisfy $\left\{\begin{array}{l}
x+3\gt 0\\
x-5\gt 0
\end{array}\right.\quad \Rightarrow x\gt $5$\qquad (*)$
in order for the equation to be defined.
On the RHS, apply$ \quad \log_{4}4^{2}=2$
$\log_{4}(x+3)=\log_{4}16+\log_{4}(x-5)$
... apply $\qquad \log_{a}(MN)=\log_{a}M+\log_{a}N$
$\log_{4}(x+3)=\log[16(x-5)]$
... apply the principle of logarithmic equality
$x+3=16(x-5)$
$ x+3=16x-80\qquad$... add $80-x$
$83=15x$
$x=\displaystyle \frac{83}{15}\quad $... satisfies (*), and is a valid solution.
