Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.7 - Page 315: 23

Answer

$104.8$ million

Work Step by Step

We can model the population growth using exponential growth. If the population \(P\) at time \(t=0\) is \(P_0 = 50\) million, and it grows at a steady rate of \(3\%\) per year, then after \(n\) years, \[ P_n = P_0 \,\bigl(1 + 0.03\bigr)^{n} = 50 \times (1.03)^n \text{ million}. \] For \(n = 25\) years, \[ P_{25} = 50 \times (1.03)^{25} \text{ million}. \] If you wish to approximate numerically: \[ (1.03)^{25} \approx 2.096 \quad\Longrightarrow\quad P_{25} \approx 50 \times 2.096 = 104.8 \text{ million}. \]
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