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: 7


a. $A=6.04e^{0.01t}$ b. in 2040.

Work Step by Step

$a.$ $A=A_{0}e^{kt }$, and we are given: $ A_{0 }=6.04\quad$ (for t=0 in 2000) $A=10 \quad $when t=50, in 2050, From this we find k: $10=6.04e^{50k}\qquad/\div 6.04$ $\displaystyle \frac{10}{6.04}=e^{50k}\qquad$ ... apply ln( ) to both sides $\ln \displaystyle \frac{10}{6.04} =50k\qquad /\div 50$ $k= \displaystyle \frac{\ln \frac{10}{6.04}}{50}\approx$0.0100836216209$\approx 0.01$ So, our model is $A=6.04e^{0.01t}$ b. Find t for which A= 9. $9=6.04e^{0.01t}\qquad/\div 6.04$ $\displaystyle \frac{9}{6.04}=e^{0.01t}\qquad$ ... apply ln( ) to both sides $\displaystyle \ln \frac{9}{6.04}=0.01t\qquad /\times 100$ $ t=100\displaystyle \ln \frac{9}{6.04}\approx$39.8820565389$\approx 40$ 40 years after 2000 is the year 2040.
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