College Algebra (11th Edition)

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

Chapter 4 - Review Exercises - Page 469: 48



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 100(1.02)^{s/4}=200 ,$ divide both sides by $100.$ Then take the logarithm of both sides. Use the properties of logarithms and of equality to isolate the variable. Finally, express the answer with $3$ decimal places. $\bf{\text{Solution Details:}}$ Dividing both sides by $100,$ the equation above is equivalent to \begin{array}{l}\require{cancel} (1.02)^{s/4}=\dfrac{200}{100} \\\\ (1.02)^{s/4}=2 .\end{array} Taking the logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \log(1.02)^{s/4}=\log2 .\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{s}{4}\log(1.02)=\log2 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 4\left( \dfrac{s}{4}\log(1.02) \right)=4(\log2) \\\\ s\log(1.02)=4\log2 \\\\ s=\dfrac{4\log2}{\log(1.02)} \\\\ s\approx140.011 .\end{array}
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