The logarithm of $27\sqrt[4]{9}\sqrt[3]{9}$ to the base $3$ is: 
$\textbf{(A)}\ 8\frac{1}{2} \qquad \textbf{(B)}\ 4\frac{1}{6} \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \text{none of these}$
$27\sqrt[4]{9}\sqrt[3]{9}$ can be rewritten as $3^3\cdot 3^\frac{1}{2}\cdot 3^\frac{2}{3}$. Using exponent rules, this simplifies to $3^\frac{25}{6}$.
The problem wants us to find $\log_3{27\sqrt[4]{9}\sqrt[3]{9}}$. We just found that this is equal to $\log_3{3^\frac{25}{6}}$. Using logarithm rules, this is equal to $\frac{25}{6}\log_3{3}$, which is simply $\frac{25}{6}$. The answer is $\boxed{\text{(B)}\ 4\frac{1}{6}}$.