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: 88



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ m=6-2.5\log \left( \dfrac{M}{M_0} \right) ,$ in terms of $ M ,$ 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} 2.5\log \left( \dfrac{M}{M_0} \right)=6-m \\\\ \dfrac{2.5\log \left( \dfrac{M}{M_0} \right)}{2.5}=\dfrac{6-m}{2.5} \\\\ \log \left( \dfrac{M}{M_0} \right)=\dfrac{6-m}{2.5} .\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} \left( \dfrac{M}{M_0} \right)=\dfrac{6-m}{2.5} \\\\ \dfrac{M}{M_0}=10^{\frac{6-m}{2.5}} .\end{array} Using the properties of equality to isolate the needed variable results to \begin{array}{l}\require{cancel} M_0\cdot\dfrac{M}{M_0}=M_0\cdot10^{\frac{6-m}{2.5}} \\\\ M=M_0\left(10^{\frac{6-m}{2.5}}\right) .\end{array}
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