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: 85


$t=-\dfrac{2}{R}\ln \left(1-\dfrac{RI}{E} \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ I=\dfrac{E}{R}(1-e^{-Rt/2}) ,$ in terms of $ t ,$ use the properties of equality and logarithms to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{R}{E}\cdot I=\dfrac{R}{E}\cdot\dfrac{E}{R}(1-e^{-Rt/2}) \\\\ \dfrac{RI}{E}=1-e^{-Rt/2} \\\\ e^{-Rt/2}=1-\dfrac{RI}{E} .\end{array} Taking the natural logarithm of both sides results to \begin{array}{l}\require{cancel} \ln e^{-Rt/2}=\ln \left(1-\dfrac{RI}{E} \right) .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} -\dfrac{Rt}{2}\ln e=\ln \left(1-\dfrac{RI}{E} \right) .\end{array} Since $\ln e=1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} -\dfrac{Rt}{2}(1)=\ln \left(1-\dfrac{RI}{E} \right) \\\\ -\dfrac{Rt}{2}=\ln \left(1-\dfrac{RI}{E} \right) .\end{array} Using the properties of equality to isolate the needed variable results to \begin{array}{l}\require{cancel} -\dfrac{2}{R}\left(-\dfrac{Rt}{2}\right)=-\dfrac{2}{R}\ln \left(1-\dfrac{RI}{E} \right) \\\\ t=-\dfrac{2}{R}\ln \left(1-\dfrac{RI}{E} \right) .\end{array}
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