College Algebra (6th Edition)

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

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.5 - Page 505: 21


about 12.6 years

Work Step by Step

Exponential growth and decay models are given by $A=A_{0}e^{kt}$ in which $t$ represents time, $A_{0}$ is the amount present at $t=0$, and $A$ is the amount present at time $t$. If $k>0$, the model describes growth and $k$ is the growth rate. If $k<0$, the model describes decay and $k$ is the decay rate ----------- We have a decay model, and we want to find the time t in which A becomes $0.5A_{o}$ $0.5A_{o}=A_{o}e^{-0.055t} \displaystyle \qquad .../\times\frac{1}{A_{o}}$ $0.5=e^{-0.055t}\qquad .../$ take ln( ) of both sides $\ln 0.5=-0.055t\qquad .../\div(-0.055)$ $ t=\displaystyle \frac{\ln 0.5}{-0.055}\approx$12.6026760102 The half-life is about 12.6 years.
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