College Algebra (11th Edition)

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

Chapter 4 - Review Exercises - Page 469: 26

Answer

$2\log_5 x+4\log_5y+\log_5 m+\dfrac{1}{3}\log_5 p$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of logarithms to rewrite the given expression, $ \log_5 \left(x^2y^4\sqrt[5]{m^3p} \right) .$ $\bf{\text{Solution Details:}}$ 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_5 x^2+\log_5y^4+\log_5\sqrt[5]{m^3p} .\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_5 x^2+\log_5y^4+\log_5 (m^3p)^{1/5} .\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} 2\log_5 x+4\log_5y+\dfrac{1}{3}\log_5 (m^3p) .\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} 2\log_5 x+4\log_5y+\dfrac{1}{3}(\log_5 m^3+\log_5 p) .\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} 2\log_5 x+4\log_5y+\dfrac{1}{3}(3\log_5 m+\log_5 p) .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2\log_5 x+4\log_5y+\dfrac{1}{3}(3\log_5 m)+\dfrac{1}{3}(\log_5 p) \\\\= 2\log_5 x+4\log_5y+\log_5 m+\dfrac{1}{3}\log_5 p .\end{array}
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