Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Cumulative Review Exercises - Page 709: 16


$\frac{ln3}{0.0575}\approx19.1\ years$

Work Step by Step

Using the formula $A=A_0e^{rt}$ with $A=3A_0, r=0.0575$, we have $e^{0.0575t}=3$; thus $t=\frac{ln3}{0.0575}\approx19.1\ years$
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