Answer
The solution set is $\{4\}$.
Work Step by Step
The given equation is
$\log (x+3) + \log (x-2) = \log 14$
Use product rule.
$\log (x+3)(x-2) = \log 14$
$ (x+3)(x-2) = 14$
Use the distributive property.
$x^2+3x-2x-6=14$
Subtract $14$ from both sides.
$x^2+x-6-14=14-14$
Simplify.
$x^2+x-20=0$
Factor.
$x^2+5x-4x-20=0$
$x(x+5)-4(x+5)=0$
$(x+5)(x-4)=0$
Set both factors equal to zero.
$x+5=0$ or $x-4=0$
Isolate $x$.
$x=-5$ or $x=4$.
$-5$ does not exist in the logarithmic function domain.
Hence, the solution is $x=4$.
