Answer
$x=1$
Work Step by Step
First, the solutions must satisfy $\left\{\begin{array}{l}
x+5\gt 0\\
x+1\gt 0
\end{array}\right.\quad \Rightarrow x\gt -1\qquad (*)$
in order for the equation to be defined.
On the RHS, apply$ \quad\log_{a}(MN)=\log_{a}M+\log_{a}N$
$\ln[(x+5)(x+1)]=\ln 12$
... apply the principle of logarithmic equality
$(x+5)(x+1)=12$
$x^{2}+6x+5=12$
$ x^{2}+6x-7=0\quad$to factor, find factors of $-7$ with sum $+6$
... these are $-1$ and $+7$
$(x+7)(x-1)=0$
Possible solutions:
$ x=-7\qquad$... does not satisfy (*), not a solution.
$x=1\qquad $... satisfies (*), and is a valid solution.
