Posted by **Amie** on Wednesday, February 29, 2012 at 3:36am.

How should I approach this, what formula do I use to solve and how do I solve it?

How long, to the nearest year, will it take me to become a millionaire if I invest $3000 at 9% interest compounded continuously?

Thanks

- Math -
**Amie**, Wednesday, February 29, 2012 at 3:37am
I got the wrong answer of 12 years

- Math -
**MathMate**, Wednesday, February 29, 2012 at 10:22am
$3000 invested at continuous interest rate of 9% (=0.09) for 12 years yield a future value of

3000*e^(0.09*12)

=8834

If you want to be a millionaire, you'll need to wait a little longer!

3000*e^(0.09*n)=1000000

e^(0.09n)=3000/1000000=1000/3

Take natural log on both sides

0.09n=ln(1000/3)

n=ln(1000/3)/0.09=64.55

=approx. 64 years & 5 months and a half

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