Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 5 - 5.5 - Exponential and Logarithmic Models - 5.5 Exercises - Page 407: 43b


13th Month

Work Step by Step

We have $p(t)=\dfrac{1000}{1+9e^{-0.1656t}}$ Plug in $p(t)=500$ Therefore, $500=\dfrac{1000}{1+9e^{-0.1656}}$ or, $2= 1+9e^{-0.1656t}$ or, $e^{-0.1656t} =\dfrac{1}{9}$ or, $\ln e^{-0.1656t} =\ln \dfrac{1}{9}$ or, $t \approx 13$ months
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