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: 81

Answer

$x=e^{\frac{k}{p-a}}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ p=a+\dfrac{k}{\ln x} ,$ for $ x ,$ 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 $ x ,$ the equation above is equivalent to \begin{array}{l}\require{cancel} p-a=\dfrac{k}{\ln x} \\\\ \ln x(p-a)=\left( \dfrac{k}{\ln x} \right) \ln x \\\\ \ln x(p-a)=k \\\\ \ln x=\dfrac{k}{p-a} .\end{array} Since $\ln x=\log_e x$, the equation above is equivalent to \begin{array}{l}\require{cancel} \log_e x=\dfrac{k}{p-a} .\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} x=e^{\frac{k}{p-a}} .\end{array}
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