Precalculus (6th Edition) Blitzer

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

Chapter 3 - Test - Page 515: 25


$\approx 6.9\%$

Work Step by Step

Re-arrange as: $2=e^{10 r}$ We need to take $\ln $ of each side of the equation. $\ln 2 =\ln e^{10 r}$ Use the logarithmic property $\ln e^a =a $ with $ a=10r $ to obtain $\ln 2= 10r \implies r = \dfrac{\ln 2}{10} \approx 0.069$ or, $\approx 6.9\%$ We can see that the money will be doubled in 10 years with an interest rate of $\approx 6.9\%$ .
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