## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{1}{5}(b+a)$
$\ln \sqrt[5]{6} = \ln 6^{\frac{1}{5}}$ $\because \log_a M^r = r \log_a M$ $\therefore \ln 6^{\frac{1}{5}} = \frac{1}{5} \ln{6}$ $\because \log_a (MN) = \log_a M+\log_a N$ $\therefore \ln6 = \ln(3 \cdot 2)= \ln3+\ln2$ With $\ln3 = b \hspace{20pt} \text{and} \hspace{20pt} \ln2=a$, then $\ln3+\ln2 = b+a$ $\ln6 = b+a$ Therfeore, $\ln \sqrt[5]{6} =\frac{[1}{5}\ln{6} = \boxed{\dfrac{1}{5}(b+a)}$