College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 405: 90

Answer

a) $7328.73$ b) $3.46$ years

Work Step by Step

In $A(t)=Pe^{rt}$ for continuous interest, $P,r,t$ respectively stand for the principal, interest rate per year and the number of years. $A(t)$ is the amount after $t$ years. So if we invest $P=6500$ at an interest rate of $r=0.06$, the amount after $t$ years is: a) $A(2)=6500e^{0.06(2)}\approx7328.73$ b) In this case: $A=8000$. Then our equation is: $8000=6500e^{0.06(t)}\\16/13=e^{0.06(t)}\\0.06t=\ln(16/13)\\t=\frac{\ln(16/13)}{0.06}\approx3.46$ years
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