Evaluate $\log_{\sqrt{6}} (216\sqrt{6})$.
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, $x=\boxed{7}$.
7
Level 3
Algebra
Problem: Evaluate $\log_{\sqrt{6}} (216\sqrt{6})$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To evaluate $\log_{\sqrt{6}} (216\sqrt{6})$, we first express $216\sqrt{6}$ in terms of base $6$. 

We know that $216 = 6^3$. Therefore, we can rewrite $216\sqrt{6}$ as $6^3 \cdot 6^{1/2}$. 

Using the properties of exponents, we combine the terms: 
$6^3 \cdot 6^{1/2} = 6^{3 + 1/2} = 6^{3.5} = 6^{7/2}$.

Next, we apply the change of base formula for logarithms:
$\log_{\sqrt{6}} (6^{7/2})$ can be simplified using the property $\log_b (a^c) = c \cdot \log_b (a)$.

Thus, we have:
$\log_{\sqrt{6}} (6^{7/2}) = \frac{7}{2} \cdot \log_{\sqrt{6}} (6)$.

Now, we need to evaluate $\log_{\sqrt{6}} (6)$. We can express $\sqrt{6}$ as $6^{1/2}$, leading to:
$\log_{\sqrt{6}} (6) = \frac{\log_6 (6)}{\log_6 (\sqrt{6})} = \frac{1}{1/2} = 2$.

Substituting this back into our earlier expression gives:
$\frac{7}{2} \cdot 2 = 7$.

Thus, the final answer is $7$.<|END_OF_LATENT|> The final answer is $7$.