Answer
Solution set = $\{8\}$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{lllll}
x-3 \gt 0, & and & x \gt 0 & and & x+2 \gt 0\\
x \gt 3 & & x \gt 0 & & x \gt -2
\end{array}\right]$
$ x \gt $3 $\qquad (*)$
This is the condition eventual solutions must satisfy.
$\log_{2}(x-3)+\log_{2}x-\log_{2}(x+2)=2$
LHS: apply the product rule; RHS write $2$ as $\log_{2}2^{2}$
$\log_{6}[(x-3)x]-\log_{2}(x+2)=\log_{2}2^{2}$
LHS: apply the quotient rule,
$\displaystyle \log_{6}\frac{x(x-3)}{x+2}=\log_{2}4$
If $\log_{b}M=\log_{b}N,$ then M=N
$\displaystyle \frac{x(x-3)}{x+2}=4\quad $ ... multiply with $(x+2)$
$ x(x-3)=4(x+2)$
$ x^{2}-3x= 4x+8$
$ x^{2}-7x-8=0$
Factor: -8 and +1 are factors of -8 whose sum is -7
$(x+1)(x-8)=0$
$ x=-1$ - does not satisfy (*), discard
$ x=8$ - satisfies (*)
Thus, the solution set = $\{8\}$
