Answer
Solution set: $ \{-2+4\sqrt{2}\}$ or $\{3.6569\}$
Work Step by Step
(*) For the equation to be defined, it must be true that
$\left\{x+2 \gt 0 \right.$
Thus, $x \gt -2 $
...Apply$: \quad \log_{a}M^{r}=r\log_{a}M\quad $
$\log_{6}(x+2)^{2}=\log_{6}2^{3}+\log_{6}4$
...Apply$: \quad \log_{a}(MN)=\log_{a}M+\log_{a}N$
$\log_{6}(x+2)^{2}=\log_{6}(2^{3}\cdot 4)$
...Apply$: \quad $If $\log_{a}M=\log_{a}N,$ then $M=N$
$(x+2)^{2}=32$
$x+2=\pm\sqrt{2^{5}}$
$x=-2\pm 4\sqrt{2}$
The solutions need to satisfy (*), so $ x=-2-4\sqrt{2}\approx-7.6569$ is eliminated.
Solution set: $ \{-2+4\sqrt{2}\}$ or $\{3.6569\}$