Answer
Solution set = $\{2\}$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{lll}
x+2 \gt 0, & and & x-1 \gt 0\\
x \gt -2 & & x \gt 1
\end{array}\right]$
$ x \gt 1 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log_{4}(x+2)-\log_{4}(x-1)=1$
LHS: apply the quotient rule; RHS: write $1$ as $\log_{4}4$
$\displaystyle \log_{4}\frac{(x+2)}{(x-1)}=\log_{4}4\qquad $
(if $\log_{b}M=\log_{b}N,$ then M=N)
$\displaystyle \frac{x+2}{x-1}=4\qquad $ ... multiply with $(x-1)$
$ x+2=4(x-1)$
$ x+2=4x-4\qquad $ ... add $4-x $
$6=3x $
$ x=2\qquad $... satisfies (*)
Solution set = $\{2\}$
