# Interest rate

• November 22nd 2009, 01:32 PM
VNVeteran
Interest rate
An initial investment of \$4000 doubles in value in 6.8 years. Assuming continuous compounding, what was the interest rate?
• November 22nd 2009, 01:40 PM
pickslides
I would use the model

$A=Pe^{rt}$

where A = annuities, P = principal amount invested, r = rate, t = time in years

Therefore

$A=4000e^{rt}$

as the amount doubled over 6.8 years, using (t,A) = (6.8,8000) gives

$8000=4000e^{6.8\times r}$

Can you solve for r?

Spoiler:
$r = \frac{\ln(2)}{6.8}\approx 10\%$
• November 22nd 2009, 01:52 PM
VNVeteran
Quote:

Originally Posted by pickslides
I would use the model

$A=Pe^{rt}$

where A = annuities, P = principal amount invested, r = rate, t = time in years

Therefore

$A=4000e^{rt}$

as the amount doubled over 6.8 years, using (t,A) = (6.8,8000) gives

$8000=4000e^{6.8\times r}$

Can you solve for r?

Spoiler:
$r = \frac{\ln(2)}{6.8}\approx 10\%$

10.19% just what i was looking for