## Precalculus (6th Edition) Blitzer

False. $$\frac{\log_28}{\log_24}=\frac{3}{2}$$
As we know, $\log_ab^c=c\log_ab$. So we have$$\log_24=\log_22^2=2\log_22=2(1)=2 \\ \log_28=\log_22^3=3\log_22=3(1)=3$$Thus,$$\frac{\log_28}{\log_24}=\frac{3}{2}$$