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 411: 48

Answer

$41$ years

Work Step by Step

The half life for carbon is 5730 years. so, $C=C_0e^{kt}$ $(1/2)C_0=C_0e^{5700k} \implies k=-0.0001216$ Now, $0.995 C_0=C_0e^{-0.0001216t}$ $C=C_0e^{-\dfrac{\ln 2}{5730}(5000)}$ s$\implies \ln (0.995)=-0.0001216 t$ Thus, $t \approx 41$ years
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