smc

$27\sqrt[4]{9}\sqrt[3]{9}$ 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 $\boxed{4\frac{1}{6}}$.