Let $a=\log _{4} 9$ and $b=108 \log _{3} 8$. What is the integer equal to $\sqrt{a b}$ ?
In this solution, "log" (without explicitly indicating the base) means " $\log _{10} "$. Using logarithm rules, we have

$$
\begin{aligned}
a b & =\left(\log _{4} 9\right)\left(108 \log _{3} 8\right) \\
& =108 \frac{\log 9}{\log 4} \times \frac{\log 8}{\log 3} \\
& =108 \frac{\log 3^{2}}{\log 2^{2}} \times \frac{\log 2^{3}}{\log 3} \\
& =\frac{108 \times 2 \times 3 \times(\log 3) \times(\log 2)}{2 \times(\log 2) \times(\log 3)} \\
& =108 \times 3 \\
& =324
\end{aligned}
$$

Since $a b=324=18^{2}$ and $\sqrt{a b}>0$, we have $\sqrt{a b}=18$.

ANSWER: 18