Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 1 - Ingredients of Change: Functions and Limits - 1.1 Activities - Page 10: 25

Answer

$t(n)=7.5$ $then$ $n≈1.386$ $t(n)=1.8$ $then$ $n≈-2.599$

Work Step by Step

Solving the equation $t(n)=y$, then we have: $t(n)=y ⇔ \frac{15}{1+2e^{-0.5n}}=y ⇔ \frac{1+2e^{-0.5n}}{15}=\frac{1}{y}⇔$ $⇔1+2e^{-0.5n}=\frac{15}{y}⇔2e^{-0.5n}=\frac{15}{y}-1⇔$ $⇔e^{-0.5n}=(\frac{15}{y}-1):2⇔-0.5n=\ln((\frac{15}{y}-1):2)⇔$ $⇔n=\frac{\ln((\frac{15}{y}-1):2)}{(-0.5)}$ For $y=7.5$ $then$ $n=1.386$ and for $y=1.8$ $then$ $n=-2.599$
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