Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.4 Logarithmic Functions - 5.4 Assess Your Understanding - Page 297: 125


$h=2.29$ hours

Work Step by Step

$D=5e^{-0.4h}$, hence if $D=2$, then $2=5e^{-0.4h}$. Solve the equation above to obtain: \begin{align*}\frac{2}{5}&=e^{-0.4h}\\ \ln{\frac{2}{5}}&=-0.4h\\ \dfrac{\ln{\frac{2}{5}}}{-0.4}&=h\\ 2.29&\approx h\end{align*}.
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