## Precalculus (10th Edition)

$1$
Recall: $\log _{a}m+\log _{a}n=\log _{a}\left( m\times n\right)$ Thus, $\log _{8}2+\log _{8}4=\log _{8}\left( 2\times 4\right) =\log _{8}8$ Since $\log_a{a}=1$, then $\log_8{8}=1$