Asked by Ericka
What interest rate, compounded monthly, would you need to be paid in order to grow a $2000 investment into $3500 over a five year period?
Answers
Answered by
MathMate
Use the compound interest formula for
n=number of periods = 5*12 = 60
R=monthly interest rate, 1.01 for 1%
P=present value of investment = $2000
F=future value of investment = $3500
Then
F=AR<sup>n</sup>
3500=2000R<sup>60</sup>
rearrange
R<sup>n</sup> = 3500/2000 = 1.75
Take log on both sides
60*log(R)=log(1.75)
log(R)=log(1.75)/60
R = e<sup>log(1.75)/60</sup>
=1.00937
The equivalent annual interest. r
=(R-1)*12=11.24%
Note: in the above expression, I used log(x) for natural log, usually written as ln(x). You would have got the same answer using log to the base 10, but at the end, the values must be raised to the power of 10 (instead of e).
n=number of periods = 5*12 = 60
R=monthly interest rate, 1.01 for 1%
P=present value of investment = $2000
F=future value of investment = $3500
Then
F=AR<sup>n</sup>
3500=2000R<sup>60</sup>
rearrange
R<sup>n</sup> = 3500/2000 = 1.75
Take log on both sides
60*log(R)=log(1.75)
log(R)=log(1.75)/60
R = e<sup>log(1.75)/60</sup>
=1.00937
The equivalent annual interest. r
=(R-1)*12=11.24%
Note: in the above expression, I used log(x) for natural log, usually written as ln(x). You would have got the same answer using log to the base 10, but at the end, the values must be raised to the power of 10 (instead of e).
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