## College Algebra (11th Edition)

$\dfrac{1}{3}\log_4{a}+\dfrac{1}{4}\log_4{b} -\dfrac{1}{2}\log_4 {c}-\dfrac{2}{3}\log_4d$
$\bf{\text{Solution Outline:}}$ 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}