jmc

Let $x={\log }_{\sqrt{6}}(216\sqrt{6})$. Putting this in exponential notation gives $(\sqrt{6}{)}^x=216\sqrt{6}$. Writing both sides with $6$ as the base gives us $6^{\frac{x}{2}}=6^3\cdot 6^{\frac{1}{2}}=6^{\frac{7}{2}}$, so $x/2=7/2$. Therefore, $x=\boxed{7}$.