Answer
Solution set = $\{-3\}$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{lll}
x+6 \gt 0, & and & x+4 \gt 0\\
x \gt -6 & & x \gt -4
\end{array}\right]$
$ x \gt -4 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log_{3}(x+6)+\log_{3}(x+4)=2$
LHS: apply the product rule; RHS write $2$ as $\log_{3}3^{2}$
$\log_{5}[(x+6)(x+4)]=\log_{3}3^{2}\qquad $
(if $\log_{b}M=\log_{b}N,$ then M=N)
$(x+6)(x+4)=3^{2}$
$ x^{2}+10x+24=9$
$ x^{2}+10x+15=0$
Factor, by finding two factors of $15$ whose sum is $10$
$(x+5)(x+3)=0$
$ x=-5$ or $ x=-3$
The solution $ x=-5$ does not satisfy (*), so the
solution set = $\{-3\}$
