Answer
$4v+\dfrac{1}{2}u$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the properties of logarithms to express the given expression, $
\ln(b^4\sqrt{a})
,$ 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 Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\ln b^4+\ln\sqrt{a}
.\end{array}
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 b^4+\ln a^{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}
4\ln b+\dfrac{1}{2}\ln a
.\end{array}
Substituting $u$ with $\ln a$ and $v$ with $\ln b,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
4v+\dfrac{1}{2}u
.\end{array}
