College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 5 - Summary, Review, and Test - Cumulative Review Exercises (Chapters 1-5) - Page 588: 35

Answer

$\dfrac{\ln 3}{10}\approx 11\%$

Work Step by Step

We are given: $A=18000$ $P=6000$ $t=10$ Determine the rate $r$ using the formula: $A=Pe^{rt}$ $18000=6000e^{10r}$ $\dfrac{18000}{6000}=e^{10r}$ $3=e^{10r}$ $\ln 3=\ln e^{10r}$ $\ln 3=10r$ $r=\dfrac{\ln 3}{10}\approx 0.11=11\%$
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