Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Section 4.6 - Modeling with Exponential Functions - 4.6 Exercises - Page 379: 11

Answer

(a) $n_0=20000$ (b) $n(t)=20000e^{0.1096t}$ (c) $48064$ (d) $14.7$ years

Work Step by Step

(a) Since 2010 corresponds to $t=0$, we can read from the graph that $n_0=20000$ (b) Write the exponential model in a general form $n(t)=n_0e^{rt}$, given $n_0=20000, n(4)=31000$ from the graph, we have $31000=20000e^{4r}$ which gives $r=ln(31/20)/4=0.1096$, thus the model can be written as $n(t)=20000e^{0.1096t}$ (c) Since 2018 corresponds to $t=8$, we have $n(8)=20000e^{0.1096\times8}=\approx48064$ (d) Let $n(t)=100000=20000e^{0.1096t}$, we can obtain $t=ln(10/2)/0.1096\approx14.7$ years
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