Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Section 3.5 - Exponential Growth and Decay; Modeling Data - Exercise Set - Page 506: 27


a. See explanations. b. $0.1069$ billion years or $106,900,000$ years.

Work Step by Step

a. Assume the model function is $A=A_0e^{-kt}$ For $t=1.31$ billion years, we have $A=\frac{A_0}{2}$ Thus $\frac{A_0}{2}=A_0e^{-1.31k}$ and $k=-\frac{ln(1/2)}{1.31}\approx0.52912$ So we have the model function as $A=A_0e^{-0.52912t}$. b. Letting $A=0.945A_0$, we have $0.945A_0=A_0e^{-0.52912t}$; thus $t=-\frac{ln(0.945)}{0.52912}\approx0.1069$ billion years or $106,900,000$ years.
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