## Calculus (3rd Edition)

Using the properties of $\ln (u/v)=\ln u-\ln v$, $e^{ln u}=u$, we have (a) $\ln (a/b)=\ln a-\ln b$, so (a) matches (ii) (b) $\ln a/\ln b$ has no special forms, so (b) does not match. (c) $e^{(\ln a-\ln b )}=e^{\ln (a/ b) }=a/b$, so (c) matches (iii). (d) $(\ln a)( \ln b )$ has no special forms, so (d) does not match.