College Algebra (11th Edition)

$1+\dfrac{1}{2}\log_23-\log_25$
$\bf{\text{Solution Outline:}}$ 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}