## Calculus 8th Edition

(a) $2$ (b) $-3$
(a) Use logarithmic base formulas $\log_{b}x=\frac{log x}{log b}$ $\log_{1.5}2.25=\frac{log (2.25)}{log (1.5)}$ Thus, $\frac{log (2.25)}{log (1.5)}=\frac{log (1.5)^{2}}{log (1.5)}=2$ (b) Use logarithmic properties, $logx+logy=log(xy)$, $y logx=logx^{y}$ and $log x-log y=log(\frac{x}{y})$ $\log_{5}4-\log_{5}500=log_{5}(\frac{4}{500})=log_{5}(5^{-3})=-3$