## Precalculus (6th Edition) Blitzer

Let $\log_bM=x$ and $\log_bN=y$. So we have$$M=b^x, \quad N=b^y \quad \Rightarrow \quad \frac{M}{N}=\frac{b^x}{b^y}=b^{x-y} \\ \quad \Rightarrow \quad \log_b \left ( \frac{M}{N} \right )=x-y=\log_bM- \log_bN$$