## Precalculus (6th Edition) Blitzer

Please note that we have the following relation:$$\log_b \left ( \frac{M}{N} \right )= \log_bM - \log_bN$$However, we have$$\frac{\log_749}{\log_77}=\frac{\log_77^2}{\log_77}=2$$But,$$\log_749-\log_77= \log_7 \left ( \frac{49}{7} \right ) =\log_77=1$$