Answer
$\dfrac{1}{2}\log_m5+\dfrac{3}{2}\log_mr-\dfrac{5}{2}\log_mz$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}