jmc

Let $x={\log }_{4}32$. Then, we must have $4^x=32$. Writing both 4 and 32 with 2 as the base gives $(2^2{)}^x=2^5$, so $2^{2x}=2^5$. Therefore, we must have $2x=5$, so $x=\boxed{\frac{5}{2}}$.