## Precalculus (10th Edition)

$1$
We know that $\log_a {x}-\log_a {y}=\log_a {\left(\frac{x}{y}\right)}$. Hence, $\log_8 {16}-\log_8 {2} \\=\log_8{\left(\frac{16}{2}\right)}\\ =\log_8 {8}\\ =\log_8 {8^1}$. We know that $\log_a {a^x}=x$. Thus, $\log_8 {8^1}=1.$