Answer:
- 3 times per year: 6.416 years
- continuously: 6.301 years
Explanation:
You want the doubling time for an investment at 11% annual interest, compounded 3 times per year, and if it were compounded continuously.
Compound interest
For interest compounded n times per year at annual rate r for t years, the multiplier of the investment is ...
k = (1 +r/n)^(nt)
We want to find t when the multiplier is 2.
2 = (1 +0.11/3)^(3t)
Taking logarithms, we have ...
log(2) = 3t·log(1 +0.11/3)
t = log(2)/(3·log(1 +0.11/3)) ≈ 6.416 . . . . years
It will take 6.416 years to double.
Continuously compounded interest
When the interest is compounded continuously, the multiplier is ...
k = e^(rt)
Filling in the values of k and r, and taking logarithms, we have ...
2 = e^(0.11t)
ln(2) = 0.11t
ln(2)/0.11 ≈ 6.301 . . . . years
It would only take 6.301 years.