College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Review Exercises: 66

Answer

$n=a\left( e^{\frac{S}{a}}-1 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ S=a\ln\left(1+\dfrac{n}{a}\right) ,$ for $ n ,$ use the properties of equality to isolate the $\ln$ expression. Then change to exponential form. Use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Dividing both sides by $a,$ the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{S}{a}=\dfrac{a\ln\left(1+\dfrac{n}{a}\right)}{a} \\\\ \dfrac{S}{a}=\ln\left(1+\dfrac{n}{a}\right) .\end{array} Since $\ln x =\log_e x,$ the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{S}{a}=\log_e\left(1+\dfrac{n}{a}\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} e^{\frac{S}{a}}=1+\dfrac{n}{a} .\end{array} Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} a\left(e^{\frac{S}{a}}\right)=a\left( 1+\dfrac{n}{a} \right) \\\\ ae^{\frac{S}{a}}=a+n \\\\ ae^{\frac{S}{a}}-a=n \\\\ n=ae^{\frac{S}{a}}-a \\\\ n=a\left( e^{\frac{S}{a}}-1 \right) .\end{array}
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