## Intermediate Algebra for College Students (7th Edition)

$\frac{1}{5}\log{x} - \frac{1}{5}\log{y}$
Note that $\sqrt[5]{\frac{x}{y}} = \left(\frac{x}{y}\right)^{\frac{1}{5}}$ therefore the expression above is equivalent to: $=\log{(\frac{x}{y})^{\frac{1}{5}}}$ RECALL: (1) $\log{(b^c)}=c \cdot \log{b}$ (2) $\log{(xy)} = \log{x} + \log{y}$ (3) $\log{(\frac{x}{y})}=\log{x} - \log{y}$ Use rule (1) above to obtain: $=\frac{1}{5}\log{(\frac{x}{y})}$ Use rule (3) above to obtain: $=\frac{1}{5}(\log{x} - \log{y}) \\=\frac{1}{5}\log{x} - \frac{1}{5}\log{y}$