College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.3 - Logarithmic Functions - 4.3 Exercises: 65

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}
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