Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Cumulative Review - Page 670: 34

Answer

$\text{approximately }34.7\text{ years}$

Work Step by Step

Since the amount needs to double, then $A=10,000$. Substituting the known values in the formula $A=P\left( 1+\dfrac{r}{n} \right)^{nt}$, then, \begin{array}{l} 10,000=5,000\left( 1+\dfrac{0.02}{4} \right)^{4t} \\\\ 2=\left( 1+\dfrac{0.02}{4} \right)^{4t} \text{...divide both sides by $5,000$} \\\\ 2=\left( 1+0.005 \right)^{4t} \\ 2=\left( 1.005 \right)^{4t} \\ \log2=\log\left( 1.005 \right)^{4t} \text{...get the logarithm of both sides} \\ \log2=4t(\log1.005) \\ \dfrac{\log2}{4(\log1.005)}=t \\ t\approx34.7 .\end{array} Hence, the money will double in $ \text{approximately }34.7\text{ years} .$
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