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: 82

Answer

$t=e^{\frac{p-r}{k}}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ r=p-k\ln t ,$ for $ t ,$ use the properties of equality to isolate the needed variable. Then change to exponential form. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate $ t ,$ the equation above is equivalent to \begin{array}{l}\require{cancel} k\ln t=p-r \\\\ \dfrac{k\ln t}{k}=\dfrac{p-r}{k} \\\\ \ln t=\dfrac{p-r}{k} .\end{array} Since $\ln x=\log_e x$, the equation above is equivalent to \begin{array}{l}\require{cancel} \log_e t=\dfrac{p-r}{k} .\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} t=e^{\frac{p-r}{k}} .\end{array}
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