## College Algebra (11th Edition)

$\dfrac{1}{2}\log_m5+\dfrac{3}{2}\log_mr-\dfrac{5}{2}\log_mz$
$\bf{\text{Solution Outline:}}$ Use the properties of radicals and logarithms to rewrite the given expression, $\log_m\sqrt{\dfrac{5r^3}{z^5}} .$ $\bf{\text{Solution Details:}}$ Using the Quotient Rule of radicals which is given by $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}{},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_m\dfrac{\sqrt{5r^3}}{\sqrt{z^5}} .\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} \log_m\sqrt{5r^3}-\log_m\sqrt{z^5} .\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_m\sqrt[2]{5r^3}-\log_m\sqrt[2]{z^5} \\\\= \log_m(5r^3)^{1/2}-\log_mz^{5/2} .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_m\left(5^{1/2}r^{3\cdot{\frac{1}{2}}}\right)-\log_mz^{5/2} \\\\= \log_m\left(5^{\frac{1}{2}}r^{{\frac{3}{2}}}\right)-\log_mz^{5/2} .\end{array} 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} \log_m5^{\frac{1}{2}}+\log_mr^{{\frac{3}{2}}}-\log_mz^{5/2} .\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_m5+\dfrac{3}{2}\log_mr-\dfrac{5}{2}\log_mz .\end{array}