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 - Page 424: 92



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Laws of Logarithms to find an equivalent expression for the given expression, $ \log_{10} 36^{1/3} ,$ which uses the given logarithmic values, $\log_{10} 2= 0.3010 $ and/or $\log_{10} 3= 0.4771 .$ $\bf{\text{Solution Details:}}$ The given expression can be expressed as \begin{array}{l}\require{cancel} \log_{10} (4\cdot9)^{1/3} \\\\= \log_{10} (2^2\cdot3^2)^{1/3} .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_{10} 2^{2\cdot\frac{1}{3}}\cdot3^{2\cdot\frac{1}{3}} \\\\= \log_{10} 2^{\frac{2}{3}}\cdot3^{\frac{2}{3}} .\end{array} Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the given expression is equivalent \begin{array}{l}\require{cancel} \log_{10} 2^{\frac{2}{3}}+\log_{10}3^{\frac{2}{3}} .\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} \dfrac{2}{3}\log_{10} 2+\dfrac{2}{3}\log_{10}3 \\\\= \dfrac{2}{3}(\log_{10} 2+\log_{10}3) .\end{array} Substituting the known values of the logarithmic expressions results to \begin{array}{l}\require{cancel} \dfrac{2}{3}(0.3010+0.4771) \\\\= \dfrac{2}{3}(0.7781) \\\\= 0.5187 .\end{array}
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