Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.5 Exponential Growth and Decay - 6.5 Exercises - Page 472: 11

Answer

$\approx$ $2500$ years

Work Step by Step

Let $y(t)$ be the level of radioactivity So $y(t)$ = $y(0)e^{-kt}$ and $k$ is determined by using the half-life $y(5730)$ = $\frac{1}{2}y(0)$ $y(0)e^{-5730k}$ = $\frac{1}{2}y(0)$ $e^{-5730k}$ = $\frac{1}{2}$ $k$ = $\frac{\ln2}{5730}$ If $74$% of the C remains, then we know that $y(t)$ = $0.74y(0)$ $0.74$ = $e^{\frac{-t(\ln2)}{5730}}$ $t$ $\approx$ $2489$ years $\approx$ $2500$ years
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