Answer
$1+\dfrac{1}{2}\log_57-\log_53$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the properties of logarithms to rewrite the given expression, $
\log_5\dfrac{5\sqrt{7}}{3}
.$
$\bf{\text{Solution Details:}}$
Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\log_5(5\sqrt{7})-\log_53
.\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_55+\log_5\sqrt{7}-\log_53
.\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_55+\log_5\sqrt[2]{7}-\log_53
\\\\=
\log_55+\log_57^{1/2}-\log_53
.\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}
\log_55+\dfrac{1}{2}\log_57-\log_53
.\end{array}
Since $\log_bb=1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
1+\dfrac{1}{2}\log_57-\log_53
.\end{array}