College Algebra (6th Edition)

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

Chapter 4 - Summary, Review, and Test - Test - Page 514: 25


$ 6.9\%$

Work Step by Step

After $t$ years, the balance, $A$, in an account with principal $P$ and annual interest rate $r$ (in decimal form) is given by one of the following formulas: 1. For $n$ compoundings per year: $A=P(1+\displaystyle \frac{r}{n})^{nt}$ 2. For continuous compounding: $A=Pe^{rt}$. ----------------- We insert A=2P (the principal doubles), t=10 into formula (1), and solve for r: $2P=2e^{r\cdot 10}$ $ 2=e^{10r}\qquad$... apply ln( ) to both sides... $\ln 2=10r\qquad/\div 10$ $ r=\displaystyle \frac{\ln 2}{10}\approx$0.069314718056$\approx 6.9\%$
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