Answer
$1$
Work Step by Step
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,
$\log_8 {8^1}=1.$
