## University Calculus: Early Transcendentals (3rd Edition)

(a) $\frac{\log_9x}{\log_3x}=\frac{1}{2}$ (b) $\frac{\log_{\sqrt{10}}x}{\log_{\sqrt2}x}=\log_{10}2$ (c) $\frac{\log_ab }{\log_ba}=(\log_ab)^2$
*Recall the Change of Base Formula: $$\log_ax=\frac{\ln x}{\ln a}$$ (a) $$\frac{\log_9x}{\log_3x}$$ Apply the Change of Base Formula here: $\log_9x=\frac{\ln x}{\ln 9}$ and $\log_3x=\frac{\ln x}{\ln 3}$ Therefore, $$\frac{\log_9x}{\log_3x}=\frac{\frac{\ln x}{\ln 9}}{\frac{\ln x}{\ln3}}=\frac{\ln x}{\ln 9}\times\frac{\ln3}{\ln x}=\frac{\ln 3}{\ln9}=\frac{\ln3}{\ln3^2}$$ Apply Power Rule: $$\frac{\log_9x}{\log_3x}=\frac{\ln3}{2\ln3}=\frac{1}{2}$$ (b) $$\frac{\log_{\sqrt{10}}x}{\log_{\sqrt2}x}$$ Apply the Change of Base Formula here: $\log_{\sqrt{10}}x=\frac{\ln x}{\ln \sqrt{10}}$ and $\log_{\sqrt2}x=\frac{\ln x}{\ln \sqrt2}$ Therefore, $$\frac{\log_{\sqrt{10}}x}{\log_{\sqrt2}x}=\frac{\frac{\ln x}{\ln \sqrt{10}}}{\frac{\ln x}{\ln\sqrt2}}=\frac{\ln x}{\ln \sqrt{10}}\times\frac{\ln\sqrt2}{\ln x}=\frac{\ln \sqrt{2}}{\ln\sqrt{10}}=\frac{\ln(2)^{1/2}}{\ln(10)^{1/2}}$$ Apply Power Rule: $$\frac{\log_{\sqrt{10}}x}{\log_{\sqrt2}x}=\frac{\frac{1}{2}\ln2}{\frac{1}{2}\ln{10}}=\frac{\ln2}{\ln{10}}$$ Apply the Change of Base Formula again, $\frac{\ln2}{\ln10}=\log_{10}2$ Therefore, $$\frac{\log_{\sqrt{10}}x}{\log_{\sqrt2}x}=\log_{10}2$$ (c) $$\frac{\log_ab }{\log_ba}$$ Apply the Change of Base Formula here: $\log_ab=\frac{\ln b}{\ln a}$ and $\log_ba=\frac{\ln a}{\ln b}$ Therefore, $$\frac{\log_ab}{\log_ba}=\frac{\frac{\ln b}{\ln a}}{\frac{\ln a}{\ln b}}=\frac{\ln b}{\ln a}\times\frac{\ln b}{\ln a}=\Big(\frac{\ln b}{\ln a}\Big)^2$$ Apply the Change of Base Formula again, $\frac{\ln b}{\ln a}$ can be written into $\log_ab$ Therefore, $$\frac{\log_ab}{\log_ba}=(\log_ab)^2$$