Answer
$x=2\sqrt{6}$
Work Step by Step
$\log_{2} (x-4)+\log_{2} (x+4)=3, x>4$ and $x>-4$, so $x\in(4,\infty)$
$\log_{2} (x-4)(x+4)=3,$
$\log_{2} (x^2-16)=3,$ raising both sides to $2,$
$2^{\log_{2} {x^2-16}}=2^3,$
$x^2-16=8,$
$x^2=24,$
$x=\pm 2\sqrt{6}$
Only the solution $x=2\sqrt 6$ belongs to the domain.
