## College Algebra (11th Edition)

$\log_26+\log_2x-\log_2y$
$\bf{\text{Solution Outline:}}$ Use the properties of logarithms to rewrite the given expression, $\log_2\dfrac{6x}{y} .$ $\bf{\text{Solution Details:}}$ 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(6x)-\log_2y .\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_26+\log_2x-\log_2y .\end{array}