## College Algebra (10th Edition)

$\dfrac{a+b}{5}$
$\bf{\text{Solution Outline:}}$ To write the given expression, $\ln\sqrt[5]{6} ,$ in terms of $a$ and $b,$ where $a=\ln2$ and $b=\ln3,$ use the laws of logarithms and substitution. $\bf{\text{Solution Details:}}$ The given expression is equivalent to \begin{array}{l}\require{cancel} \ln\sqrt[5]{2\cdot3} \\\\= \ln(2\cdot3)^{1/5} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent to: \begin{array}{l}\require{cancel} \dfrac{1}{5}\ln(2\cdot3) .\end{array} Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the expression above is equivalent to: \begin{array}{l}\require{cancel} \dfrac{1}{5}(\ln2+\ln3) .\end{array} By substitution, since $a=\ln2$ and $b=\ln3,$ the expression above is equivalent to: \begin{array}{l}\require{cancel} \dfrac{1}{5}(a+b) \\\\= \dfrac{a+b}{5} .\end{array}