## Precalculus (10th Edition)

Let $M=a^y$ and $N=a^z$, which implies $\log_a{M}=y$ and $\log_a{N}=z$. Therefore $\log_a{\frac{M}{N}}=\log_a{\frac{a^y}{a^z}}=\log_a{a^{y-z}}=y-z=\log_a{M}-\log_a{N}.$
Let $M=a^y$ and $N=a^z$, which implies $\log_a{M}=y$ and $\log_a{N}=z$. Therefore $\log_a{\frac{M}{N}}=\log_a{\frac{a^y}{a^z}}=\log_a{a^{y-z}}=y-z=\log_a{M}-\log_a{N}.$