SOLUTION: Suppose $7,100 is invested in an account at an annual interest rate of 3.4% compounded continuously. How long (to the nearest tenth of a year) will it take the investment to double

Question 1040540: Suppose $7,100 is invested in an account at an annual interest rate of 3.4% compounded continuously. How long (to the nearest tenth of a year) will it take the investment to double in size? Answer: ________
Answer by Aldorozos(172) (Show Source): You can put this solution on YOUR website!
F= future value
P=present value
F=p*e*rt
Read below to understand continuous compounding
http://cs.selu.edu/~rbyrd/math/continuous/
F=2p
2p =p * e^.034t
Divide both sides by p
2= e^.034t t is the time
You have to know properties of log and Ln to solve this
Get Ln of both sides
Ln of 2 = Ln of e^.034t
Ln of 2 = .034r* Ln of e
Using your calculator you will find Ln of 2 is .693 and Ln of e is one
.693 = .034 t
T = time = years = .693/.034. = 20.4 which means it takes around 20.4 years for the money to double.
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