University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 410: 27


$ \approx 59.8$ft

Work Step by Step

Given: $L=\dfrac{L_0}{2}$ and $k=18$ Now, we have $L=L_0e^{-kx}$ ...(1) Then $\dfrac{L_0}{2}=L_0e^{-18x} \implies k=\dfrac{\ln (2)}{18} \approx 0.0385$ Now, when after another t = 14 hours, that is, $t=14+10 =24$ Then equation (1) becomes $\dfrac{L_0}{10}=L_0e^{-(0.0385)x}$ This implies that $\ln (10) =0.0385 x$ Hence, $x \approx 59.8$ft
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