Answer
Solution set = $\{4\}$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{ll}
x+5 \gt 0, & x \gt 0\\
x \gt -5 &
\end{array}\right]$
$ x \gt 0 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log_{6}(x+5)+\log_{6}x=2$
LHS: apply the product rule, RHS: write $2$ as $\log_{6}6^{2}$
$\log_{5}[(x+5)x]=\log_{6}6^{2}\qquad $ ... if $\log_{b}M=\log_{b}N,$ then M=N
$(x+5)x=6^{2}$
$ x^{2}+5x=36$
$ x^{2}+5x-36=0$
Factor, by finding two factors of -36 whose sum is 5
$(x+9)(x-4)=0$
$ x=-9$ or $ x=4$
The negative solution does not satisfy (*), so the
solution set = $\{4\}$
