## Algebra and Trigonometry 10th Edition

It is shown that: $\frac{\log_ax}{\log_{\frac{a}{b}}x}=1+\log_a\frac{1}{b}$
Use the Change-of-Base formula, the Quotient Property and the Power Property: $\frac{\log_ax}{\log_{\frac{a}{b}}x}=\frac{\frac{\log x}{\log a}}{\frac{\log x}{\log\frac{a}{b}}}=\frac{\log x}{\log a}·\frac{\log\frac{a}{b}}{\log x}=\frac{\log\frac{a}{b}}{\log a}=\frac{\log a-\log b}{\log a}=\frac{\log a}{\log a}-\frac{\log b}{\log a}=1+\frac{-\log b}{\log a}=1+\frac{\log b^{-1}}{\log a}=1+\frac{\log \frac{1}{b}}{\log a}=1+\log_a\frac{1}{b}$