College Algebra (11th Edition)

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

Chapter 4 - Test: 6

Answer

$2\log_7 x+\dfrac{1}{4}\log_7y-3\log_7z$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of logarithms to change the form of the given expression, $ \log_7\dfrac{x^2\sqrt[4]{y}}{z^3} .$ $\bf{\text{Solution Details:}}$ U sing the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log_7(x^2\sqrt[4]{y})-\log_7z^3 .\end{array} Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log_7x^2+\log_7\sqrt[4]{y}-\log_7z^3 .\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_7x^2+\log_7y^{1/4}-\log_7z^3 .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} 2\log_7 x+\dfrac{1}{4}\log_7y-3\log_7z .\end{array}
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