Thread: Interest rate

An initial investment of $4000 doubles in value in 6.8 years. Assuming continuous compounding, what was the interest rate?

I would use the model

$\displaystyle A=Pe^{rt}$

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

Therefore

$\displaystyle A=4000e^{rt}$

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

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

Can you solve for r?

Spoiler:

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

Originally Posted by pickslides

I would use the model

$\displaystyle A=Pe^{rt}$

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

Therefore

$\displaystyle A=4000e^{rt}$

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

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

Can you solve for r?

Spoiler:

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

10.19% just what i was looking for