Answer
$x=2$
Work Step by Step
First, the solutions must satisfy $\left\{\begin{array}{l}
x+3\gt 0\\
x+2\gt 0
\end{array}\right.\quad \Rightarrow x\gt -2\qquad (*)$
in order for the equation to be defined.
On the RHS, apply$ \quad\log_{a}(MN)=\log_{a}M+\log_{a}N$
$\log_{6}[(x+3)(x+2)]=\log_{6}20$
... apply the principle of logarithmic equality
$(x+3)(x+2)=20$
$x^{2}+5x+6=20$
$ x^{2}+5x-14=0\quad$to factor, find factors of $-14$ with sum $+5$
... these are $+7$ and $-2$
$(x+7)(x-2)=0$
Possible solutions:
$ x=-7\qquad$... does not satisfy (*), not a solution.
$x=2\qquad $... satisfies (*), and is a valid solution.
