## Precalculus (6th Edition) Blitzer

$$\frac{4}{5}=0.8$$
As we know, $\log_a1=0$, $\log_aa=1$, and $\log_ab^c=c\log_ab$. So we have$$\log_381=\log_33^4=4\log_33=4(1)=4, \quad \log_{\pi }1=0, \\ \log_{2\sqrt{2}}8=\log_{\sqrt{8}}(\sqrt{8})^2=2\log_{\sqrt{8}} \sqrt{8}=2(1)=2, \quad \log 0.001= \log_{10}10^{-3}=-3\log_{10}10=-3(1)=-3$$Thus,$$\frac{\log_381-\log_{\pi }1}{\log_{2\sqrt{2}}8-\log 0.001}= \frac{4-0}{2-(-3)}= \frac{4}{5}=0.8$$