Precalculus (6th Edition) Blitzer

$x=3$
Use the rule $\log (p q)=\log p+\log q$ with $p=x+5$ and $q=x-1$ to obtain $\log_{2}(x+5)(x-1)=4$ Simplify: $(x+5)(x-1)=2^4$ $x^2+4x-5 =16$ or, $(x+7)(x-3)$ So, $x=-7, 3$, but $x=-3$ is not in the domain. Thus, our result is: $x=3$