Intermediate Algebra (6th Edition)

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

Chapter 9 - Section 9.8 - Exponential and Logarithmic Equations and Problem Solving - Graphing Calculator Explorations - Page 590: 2

Answer

15.4 years or 15 years and 5 months

Work Step by Step

1000 dollars initial amount 2000 dollars desired amount 4.5% interest Compounds daily (365 days) $A=2000$ $P=1000$ $r=.045$ $n=365$ $A=P(1+(r/n))^{n*t}$ $2000=1000(1+(.045/365))^{365*t}$ $2000/1000=1000(1+(.045/365))^{365*t}/1000$ $2=(1+(.045/365))^{365*t}$ $2=(1+.00012328)^{365*t}$ $2=1.00012328^{365*t}$ $ln 2 =365* t* ln 1.00012328$ $ln2/365*ln 1.00012328 = 365*t* ln 1.00012328$ $ .693147/365* .00012327 =t$ $.693147/.04499=t$ $15.405 =t$ 15.4 years or 15 years and 5 months
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