College Algebra (11th Edition)

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

Chapter 4 - Review Exercises - Page 469: 68



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ D=200+100\log x ,$ for $ x ,$ use the properties of equality to isolate the $\log$ expression. Then change to exponential form. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the $\log$ expression, the equation above is equivalent to \begin{array}{l}\require{cancel} D-200=100\log x \\\\ \dfrac{D-200}{100}=\log x .\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} \dfrac{D-200}{100}=\log_{10} x \\\\ 10^{\frac{D-200}{100}}=x \\\\ x=10^{\frac{D-200}{100}} .\end{array}
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