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 507: 40



Work Step by Step

Let $f(x)=7$; we have $\frac{12.57}{1+4.11e^{-0.026x}}=7$ and $4.11e^{-0.026x}=\frac{12.57}{7}-1\approx0.7957$ Thus $t=-\frac{ln(0.7957/4.11)}{0.026}\approx63$, corresponding to year $1949+63=2012$
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