College Algebra (11th Edition)

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

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 448: 92



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ D=160+10\log x ,$ in terms of $ x ,$ use the properties of equality to isolate the logarithmic expression. Then change to exponential form. Finally use again the properties of equality to isolate the needed variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} D-160=10\log x \\\\ \dfrac{D-160}{10}=\dfrac{10\log x}{10} \\\\ \dfrac{D-160}{10}=\log x \\\\ \log x=\dfrac{D-160}{10} .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} \log_{10} x=\dfrac{D-160}{10} \\\\ x=10^{\frac{D-160}{10}} .\end{array}
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