Answer
$2\log_7 x+\dfrac{1}{4}\log_7y-3\log_7z$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the properties of logarithms to change the form of the given expression, $
\log_7\dfrac{x^2\sqrt[4]{y}}{z^3}
.$
$\bf{\text{Solution Details:}}$
U sing the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent
\begin{array}{l}\require{cancel}
\log_7(x^2\sqrt[4]{y})-\log_7z^3
.\end{array}
Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the equation above is equivalent
\begin{array}{l}\require{cancel}
\log_7x^2+\log_7\sqrt[4]{y}-\log_7z^3
.\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_7x^2+\log_7y^{1/4}-\log_7z^3
.\end{array}
Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent
\begin{array}{l}\require{cancel}
2\log_7 x+\dfrac{1}{4}\log_7y-3\log_7z
.\end{array}