How do you solve 2^(x+4)=1/32?

Consider that you can write: $\frac{1}{32} = \frac{1}{2} ^ 5 = {2}^{-} 5$
${2}^{x + 4} = {2}^{-} 5$
Take the ${\log}_{2}$ of both sides and remember that ${\log}_{a} {a}^{x} = x$;
${\log}_{2} \left({2}^{x + 4}\right) = {\log}_{2} \left({2}^{-} 5\right)$
$x + 4 = - 5$
$x = - 9$