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
2 answers
I got the wrong answer of 12 years
$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
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
1 answer