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: 30

Answer

$ t\approx 6.2$ hours.

Work Step by Step

Model: $ A=A_{0}e^{kt}$ Unknown: $ k $, when $ A=0.5A_{0},\ t=12$ $0.5A_{0}=A_{0}e^{k\cdot 12}\qquad... /\div A_{0}$ $0.5=e^{k\cdot 12}\qquad.../\ln(...)$ $-0.693147=12k\qquad.../\div(12)$ $ k\approx-0.057762$ So, our model is $ A=A_{0}e^{-0.057762t}$ Now we find $ t $ for $ A=0.7A_{0}$ $0.7A_{0}=A_{0}e^{-0.057762t}\qquad... /\div A_{0}$ $0.7=e^{-0.057762t}\qquad.../\ln(...)$ $\ln 0.7 =-0.057762t\qquad.../\div(-0.057762)$ $\displaystyle \frac{\ln 0.7}{-0.057762}=t $ $ t\approx 6.2$ hours.
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