jmc

We begin by writing $\frac{1}{4}$ and $16$ as powers of $2$. $$\begin{array}{rl}(2^{-2}{)}^{2x+8}& =(2^4{)}^{2x+5}\\ 2^{-4x-16}& =2^{8x+20}\\ -4x-16& =8x+20\\ x& =\boxed{-3}\end{array}$$