## Precalculus (10th Edition)

$2$
We know that $\log_a {b}=\dfrac{\log_c {b}}{\log_c {a}}$. Hence, $\log_3 {8}\cdot \log_8 {9} \\=\dfrac{\log{8}}{\log{3}}\cdot\dfrac{\log{9}}{\log{8}} \\=\dfrac{\log{9}}{\log{3}} \\=\log_3 {9} \\=\log_3 {(3^2)}$ Use the rule $\log_a{a^x}=x$ to obtain: $\log_3{(3^2)}=2$