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: 90

Answer

$-0.1303$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Laws of Logarithms to find an equivalent expression for the given expression, $ \log_{10} \dfrac{20}{27} ,$ 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} \dfrac{10\cdot2}{3^3} .\end{array} 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_{10} (10\cdot2)-\log_{10}3^3 .\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} 10+\log_{10} 2-\log_{10}3^3 .\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} \log_{10} 10+\log_{10} 2-3\log_{10}3 .\end{array} Since $\log_b b=1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 1+\log_{10} 2-3\log_{10}3 .\end{array} Substituting the known values of the logarithmic expressions results to \begin{array}{l}\require{cancel} 1+0.3010-3(0.4771) \\\\= 1+0.3010-1.4313 \\\\= -0.1303 .\end{array}
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