College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Summary, Review, and Test - Review Exercises - Page 512: 87

Answer

325 days

Work Step by Step

$$A = A_{0}e^{kt}$$ To solve this exercise, we must first find $k$ by using the knowledge that the half-life of polonium-210 is 140 days: $$0.5 = e^{k(140)}$$ $$140k = \ln 0.5 = \ln \frac{1}{2} = \ln 2^{-1} = -\ln 2$$ $$k = -\frac{\ln 2}{140}$$ Now, we can solve for $t$ when $\frac{A}{A_{0}} = 0.2$: $$0.2 = e^{-\frac{\ln 2}{140}t} = (e^{\ln 2})^{-\frac{1}{140}t} = 2^{-\frac{1}{140}t}$$ $$-\frac{t}{140} = log_{2}0.2$$ $$t = -140(\log_{2} 0.2) \approx 325$$
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