Answer
Solution set = $\{6\}$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{lll}
x+2 \gt 0, & and & x-5 \gt 0\\
x \gt -2 & & x \gt 5
\end{array}\right]$
$ x \gt 5 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log_{2}(x+2)-\log_{2}(x-5)=3$
LHS: apply the quotient rule; RHS write $3$ as $\log_{2}2^{3}$
$\displaystyle \log_{2}\frac{(x+2)}{(x-5)}=\log_{2}2^{3}\qquad $
(if $\log_{b}M=\log_{b}N,$ then M=N)
$\displaystyle \frac{x+2}{x-5}=2^{3}$
$\displaystyle \frac{x+2}{x-5}=8\qquad $ ... multiply with $(x-5)$
$ x+2=8(x-5)$
$ x+2=8x-40\qquad $ ... add $40-x $
$42=7x $
$ x=6\qquad $... satisfies (*)
Thus, solution set = $\{6\}$
