Answer
$\log_a{M}-\log_a{N}$.
Work Step by Step
$\log_a{\frac{M}{N}}=\log_a{MN^{-1}}$.
We know that $\log_a {x}+\log_a {y}=\log_a {(x\cdot y)}$.
Thus $\log_a{MN^{-1}}=\log_a{M}+\log_a{N^{-1}}$.
We know that $\log_a {x^n}=n\cdot \log_a {x}$, hence
$\log_a{M}+\log_a{N^{-1}}=\log_a{M}-\log_a{N}$.
