AMC 12 Mock Exam 3 Problem 7

We have $\log_29=2\log_23$, $\log_316=4\log_32$, $\log_425=\frac{\log 25}{\log4}=\log_25$, and $\log_58=3\log_52$. Since $\log_23\cdot\log_32=1$ and $\log_25\cdot\log_52=1$, the product is $2\cdot4\cdot3=\boxed{24}$.