Answer
$\dfrac{1}{2}(3u-5v)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the properties of logarithms to express the given expression, $
\ln \sqrt{\dfrac{a^3}{b^5}}
,$ in a form that uses $\ln a$ and $\ln b$ only. Then substitute $u$ for any instance of $\ln a$ and substitute $v$ for any instance of $\ln b.$
$\bf{\text{Solution Details:}}$
Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\ln \left(\dfrac{a^3}{b^5}\right)^{1/2}
.\end{array}
Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{1}{2}\ln \left(\dfrac{a^3}{b^5}\right)
.\end{array}
Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{1}{2}(\ln a^3-\ln b^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
\begin{array}{l}\require{cancel}
\dfrac{1}{2}(3\ln a-5\ln b)
.\end{array}
Substituting $u$ with $\ln a$ and $v$ with $\ln b,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1}{2}(3u-5v)
.\end{array}