Answer
Solution set = $\{4\}.$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{llll}
x+3 \gt 0, & and & x-2 \gt 0 & \\
x\gt-3 & & x\gt 2 &
\end{array}\right]$
$ x \gt 2 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log(x+3)+\log(x-2)=\log 14\qquad $... LHS: product rule
$\log(x^{2}+x-6)=\log 14 \quad $ ... $\log $ is one-to-one
$ x^{2}+x-6=14$
$ x^{2}+x-20=0$
Factor: +5 and -4 are factors of -20 whose sum is +1.
$(x+5)(x-4)=0$
$ x=4 \qquad $ ... satisfies (*)
$ x=-5 \quad $ ... does not satisfy (*)
Thus, the solution set = $\{4\}.$
