#### Answer

$\dfrac{5}{3}\log_pm+\dfrac{4}{3}\log_pn-\dfrac{2}{3}\log_pt$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the properties of radicals and the properties of logarithms to rewrite the given expression, $
\log_p\sqrt[3]{\dfrac{m^5n^4}{t^2}}
.$
$\bf{\text{Solution Details:}}$
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_p\dfrac{\sqrt[3]{m^5n^4}}{\sqrt[3]{t^2}}
.\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_p\sqrt[3]{m^5n^4}-\log_p\sqrt[3]{t^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_p(m^5n^4)^{\frac{1}{3}}-\log_p(t^2)^{\frac{1}{3}}
.\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_p(m^{5\cdot\frac{1}{3}}n^{4\cdot\frac{1}{3}})-\log_pt^{2\cdot\frac{1}{3}}
\\\\=
\log_p(m^{\frac{5}{3}}n^{\frac{4}{3}})-\log_pt^{\frac{2}{3}}
.\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_pm^{\frac{5}{3}}+\log_pn^{\frac{4}{3}}-\log_pt^{\frac{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{5}{3}\log_pm+\dfrac{4}{3}\log_pn-\dfrac{2}{3}\log_pt
.\end{array}