jmc

Let $x={\log }_{\sqrt{8}}(64\sqrt{8})$. In exponential form this is $64\sqrt{8}=(\sqrt{8}{)}^x$. Since $64\sqrt{8}$ can be written as $(\sqrt{8}{)}^5$, we have $(\sqrt{8}{)}^5=(\sqrt{8}{)}^x$. Therefore, $x=\boxed{5}$.