## Calculus: Early Transcendentals (2nd Edition)

$log_{\frac{1}{b}} x =-log_{ b } x$
Assume that $b\gt0$ and $b\ne1$ Using change of base rule, $log_{\frac{1}{b}}x =\frac{ln x}{ln(\frac{1}{b})}$ Using logarithmic rule, $log_{\frac{1}{b}}x =\frac{ln x}{ln 1 - ln b}$ $log_{\frac{1}{b}}x =\frac{ln x}{- ln b}$ (Since $ln 1=0$) Again using change of base rule, $log_{\frac{1}{b}}x =-log_{b}x$