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: 90



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ d=10\log \left( \dfrac{I}{I_0} \right) ,$ in terms of $ I ,$ 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} \dfrac{d}{10}=\dfrac{10\log \left( \dfrac{I}{I_0} \right)}{10} \\\\ \dfrac{d}{10}=\log \left( \dfrac{I}{I_0} \right) .\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}{10}=\log_{10} \left( \dfrac{I}{I_0} \right) \\\\ 10^{\frac{d}{10}}=\dfrac{I}{I_0} .\end{array} Using the properties of equality to isolate the needed variable results to \begin{array}{l}\require{cancel} I_0\cdot10^{\frac{d}{10}}=I_0\cdot\dfrac{I}{I_0} \\\\ I_0\left(10^{\frac{d}{10}}\right)=I \\\\ I=I_0\left(10^{\frac{d}{10}}\right) .\end{array}
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