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: 35


$15.28$ years

Work Step by Step

Given: $P=0.9 P_0$ The exponential growth can be written as: $P=P_0e^{kt}$ ...(1) This implies that $0.9 P_0=P_0e^{k} \implies k =\ln 0.9$ As per the given statement when $P=\dfrac{1}{5} P_0$ or, $P=0.2 P_0$ Equation (1) becomes: $0.2P_0=P_0e^{(\ln 0.9)t}$ This implies that $t=\dfrac{\ln (0.2)}{\ln (0.9)}$ or, $t\approx 15.28$ years
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