College Algebra (6th Edition)

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

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.1 - Page 452: 53


a. $\$ 13,116.51$ b. $\$ 13,140.67$ c. $\$ 13,157.04$ d. $\$ 13,165.31$

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: $A=Pe^{rt}$. ------------------------ a. $A=10,000(1+\displaystyle \frac{0.055}{2})^{2\cdot 5}\approx\$ 13,116.51$ b. $A=10,000(1+\displaystyle \frac{0.055}{4})^{4\cdot 5}\approx\$ 13,140.67$ c. $A=10,000(1+\displaystyle \frac{0.055}{12})^{12\cdot 5}\approx\$ 13,157.04$ d. $A=10,000e^{0.055\cdot 5}\approx\$ 13,165.31$
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