## College Algebra (11th Edition)

$\dfrac{1}{2}\log_3x+\dfrac{1}{3}\log_3y-2\log_3w-\dfrac{1}{2}\log_3z$
$\bf{\text{Solution Outline:}}$ Use the properties of radicals and the properties of logarithms to rewrite the given expression, $\log_3\dfrac{\sqrt{x}\cdot\sqrt[3]{y}}{w^2\sqrt{z}} .$ $\bf{\text{Solution Details:}}$ 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_3(\sqrt{x}\cdot\sqrt[3]{y})-\log_3(w^2\sqrt{z}) .\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_3(\sqrt{x}\cdot\sqrt[3]{y})-(\log_3w^2+\log_3\sqrt{z}) \\\\= \log_3(\sqrt{x}\cdot\sqrt[3]{y})-\log_3w^2-\log_3\sqrt{z} \\\\= \log_3\sqrt{x}+\log_3\sqrt[3]{y}-\log_3w^2-\log_3\sqrt{z} .\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_3\sqrt[2]{x}+\log_3\sqrt[3]{y}-\log_3w^2-\log_3\sqrt[2]{z} \\\\ \log_3x^{1/2}+\log_3y^{1/3}-\log_3w^2-\log_3z^{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}\log_3x+\dfrac{1}{3}\log_3y-2\log_3w-\dfrac{1}{2}\log_3z .\end{array}