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: 38

Answer

$\approx 600$ days

Work Step by Step

Given: $A=\dfrac{1}{2} A_0$ The exponential growth can be written as: $A=A_0e^{kt}$ ...(1) This implies that $\dfrac{1}{2} A_0=A_0e^{139k} \implies k =\dfrac{\ln 0.5}{139}$ or, $k\approx -0.00499$ years Equation (1) becomes: $(0.05)A_0=A_0 e^{-0.00499t}$ $\implies t=\dfrac{\ln (0.05)}{-0.00499}$ or, $t\approx 600$ days
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