Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 402: 35

Answer

$\approx 15.28$ years

Work Step by Step

Consider the exponential growth equation as follows: $P=P_0e^{kt}$ As we are given that $P=0.9 P_0$ Then, we get $0.9 P_0=P_0e^{k} \implies k =\ln 0.9$ Now, when $P=\dfrac{1}{5} P_0$ Then, we get $P=(0.2) P_0$ $P=P_0e^{kt} \implies (0.2)P_0=(P_0)e^{(\ln 0.9)t}$ Thus, $t=\dfrac{\ln (0.2)}{\ln (0.9)} \approx 15.28$ years
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