Answer
$1+\dfrac{1}{2}\log_23-\log_25$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the properties of logarithms to rewrite the given expression, $
\log_2\dfrac{2\sqrt{3}}{5}
.$
$\bf{\text{Solution Details:}}$
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_2(2\sqrt{3})-\log_25
.\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_22+\log_2\sqrt{3}-\log_25
.\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_22+\log_2\sqrt[2]{3}-\log_25
\\\\=
\log_22+\log_23^{1/2}-\log_25
.\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_22+\dfrac{1}{2}\log_23-\log_25
.\end{array}
Since $\log_bb=1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
1+\dfrac{1}{2}\log_23-\log_25
.\end{array}