Answer
$\dfrac{1}{3}\log_4{a}+\dfrac{1}{4}\log_4{b} -\dfrac{1}{2}\log_4 {c}-\dfrac{2}{3}\log_4d
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the properties of radicals and the properties of logarithms to rewrite the given expression, $
\log_4\dfrac{\sqrt[3]{a}\cdot\sqrt[4]{b}}{\sqrt{c}\cdot\sqrt[3]{d^2}}
.$
$\bf{\text{Solution Details:}}$
Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\log_4\left(\sqrt[3]{a}\cdot\sqrt[4]{b} \right)-\log_4 \left( \sqrt{c}\cdot\sqrt[3]{d^2} \right)
.\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}
\log_4\left(\sqrt[3]{a}\cdot\sqrt[4]{b} \right)-\left(\log_4 \sqrt{c}+\log_4\sqrt[3]{d^2} \right)
\\\\=
\log_4\left(\sqrt[3]{a}\cdot\sqrt[4]{b} \right)-\log_4 \sqrt{c}-\log_4\sqrt[3]{d^2}
\\\\=
\log_4\sqrt[3]{a}+\log_4\sqrt[4]{b} -\log_4 \sqrt{c}-\log_4\sqrt[3]{d^2}
.\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}
\log_4\sqrt[3]{a}+\log_4\sqrt[4]{b} -\log_4 \sqrt[2]{c}-\log_4\sqrt[3]{d^2}
\\\\=
\log_4{a^{1/3}}+\log_4{b^{1/4}} -\log_4 {c^{1/2}}-\log_4d^{2/3}
.\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}{3}\log_4{a}+\dfrac{1}{4}\log_4{b} -\dfrac{1}{2}\log_4 {c}-\dfrac{2}{3}\log_4d
.\end{array}
