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 504: 13



Work Step by Step

Let t be years after 2010. $A=A_{0}e^{kt }$, and we are given: $ A_{0 }=82.3\quad$ (for t=0 in 2010) $A=70.5 \quad $when t=$40$, in 2050. From this we find k: $82.3=70.5e^{40k}\qquad/\div 70.5$ $\displaystyle \frac{82.3}{70.5}=e^{40k}\qquad$ ... apply ln( ) to both sides $\ln \displaystyle \frac{82.3}{70.5} =40k\qquad /\div 40$ $k= \displaystyle \frac{\ln \frac{82.3}{70.5}}{40}\approx$0.00386895994662$\approx 0.0039$
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