Answer
$\log_2a+\log_2b-\log_2c-\log_2d$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the properties of radicals and the properties of logarithms to rewrite the given expression, $
\log_2\dfrac{ab}{cd}
.$
$\bf{\text{Solution Details:}}$
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(ab)-\log_2(cd)
.\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_2(ab)-(\log_2c+\log_2d)
\\\\=
\log_2(ab)-\log_2c-\log_2d
\\\\=
\log_2a+\log_2b-\log_2c-\log_2d
.\end{array}
