## Precalculus (6th Edition) Blitzer

$$\frac{2A}{B}$$
The change-of-base property states that$$\log_bM= \frac{\log_aM}{\log_ab}$$So by introducing the new base $10$, we have$$\log_79=\frac{\log 9}{\log 7}=\frac{\log 3^2}{\log 7}= \frac{2 \log 3}{\log 7}= \frac{2A}{B}$$