jmc

Let $x={\log }_{\sqrt{5}}125\sqrt{5}$. Putting this in exponential notation gives $(\sqrt{5}{)}^x=125\sqrt{5}$. Writing both sides with 5 as the base gives us $5^{\frac{x}{2}}=5^3\cdot 5^{\frac{1}{2}}=5^{\frac{7}{2}}$, so $x/2=7/2$. Therefore, $x=\boxed{7}$.