Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 369: 97

Answer

a. 7337 fish b. about 1 year and 9 months.

Work Step by Step

(a) $P(3)=\displaystyle \frac{10}{1+4e^{-0.8(3)}}=7.337$ (thousands) After 3 years there are approximately 7337 fish . (b) Solve for {\it t}. $\displaystyle \frac{10}{1+4e^{-0.8t}}=5 \qquad $equate reciprocals $\displaystyle \frac{1+4e^{-0.8t}}{10}=\frac{1}{5}\qquad .../\times 10$ $1+4e^{-0.8t}=2 \qquad .../-1$ $4e^{-0.8t}=1 \qquad .../\div 4$ $e^{-0.8t}=0.25 \qquad$ ... apply $\ln$() to both sides $-0.8t=\ln 0.25 \qquad .../\div(-0.8)$ $t=\displaystyle \frac{\ln 0.25}{-0.8}\approx 1.733$ So the population will reach 5000 fish in about $1.73$ years (0.73 years $\approx$ 9 months)
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