Answer
$x=1$
Work Step by Step
First, the solutions must satisfy $\left\{\begin{array}{l}
x+1\gt 0\\
x+2\gt 0
\end{array}\right.\quad \Rightarrow x\gt -1\qquad (*)$
in order for the equation to be defined.
On the RHS, apply$ \quad\log_{a}(MN)=\log_{a}M+\log_{a}N$
$\log_{7}[(x+1)(x+2)]=\log_{7}6$
... apply the principle of logarithmic equality
$(x+1)(x+2)=\log_{7}6$
$x^{2}+3x+2=6$
$ x^{2}+3x-4=0\quad$to factor, find factors of -4 with sum =3
... these are $+4$ and $-1$
$(x+4)(x-1)=0$
Possible solutions:
$ x=-4\qquad$... does not satisfy (*), not a solution.
$x=1\qquad $... satisfies (*), and is a valid solution.
