## Precalculus (6th Edition) Blitzer

$\displaystyle \ln\frac{\sqrt{x}}{y}$
$\displaystyle \frac{1}{2}\ln x-\ln y=\qquad$...apply the Power Rule: $\qquad \log_{b}(M^{p})=p\cdot\log_{b}\mathrm{M}$ $=\ln x^{1/2}-\ln y\qquad$...apply the Quotient Rule: $\displaystyle \qquad \log_{b}(\frac{M}{N})=\log_{b}\mathrm{M}-\log_{b}\mathrm{N}$ $=\displaystyle \ln\frac{x^{1/2}}{y}$