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: 37


$\approx 56563$ years

Work Step by Step

Given: $A_0=10$ and $A=5$ The exponential growth can be written as: $A=A_0e^{kt}$ ...(1) This implies that $5=10e^{24360k} \implies k =\dfrac{\ln 0.5}{24360}$ or, $k\approx -0.000028454$ years Equation (1) becomes: $(0.2)(10)=10 e^{-0.000028454} t$ $\implies \dfrac{\ln (0.2)}{-0.000028454}$ or, $t\approx 56563$ years
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