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 505: 8

Answer

a. $A=3.2e^{0.026t}$ b. $2040$

Work Step by Step

a. In 2000, we have $t=0$ and $A=A_0e^{kt}=A_0e^{0}=A_0$; thus $A_0=3.2$ million. In 2050, we have $t=2050-2000=50$ and $A=3.2e^{50k}$; thus $3.2e^{50k}=12$ and $k=\frac{1}{50}ln(\frac{12}{3.2})\approx0.0264$ and we have the model function as $A=3.2e^{0.026t}$ b. Letting $A=9$, we have $3.2e^{0.026t}=9$ and $t=\frac{ln(9/3.2)}{0.026}\approx40$ which corresponds to year $2040$
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