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: 10

Answer

(a) $n(t)=350000\cdot 2^{t/25}$ (b) $n(t)=350000\cdot e^{0.0277t}$ (c) see graph. (d) $63$ years

Work Step by Step

(a) Given $n_0=350000,n(25)=2n_0$, we have $2n_0=n_02^{25/a}$ which gives $a=25$ thus the exponential model is $n(t)=350000\cdot 2^{t/25}$ (b) For this case, $2n_0=n_0e^{25r}$ and we can find $r=ln2/25=0.0277$ thus this exponential model can be written as $n(t)=350000\cdot e^{0.0277t}$ (c) The functions defined above can be graphed as shown in the figure. (d) Let $n(t)=2000000$, we have $350000e^{0.0277t}=2000000$ and we can find that $t=ln(200/35)/0.0277=62.9\approx63$ years
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