Question

Logan invested $3,800 in an account paying an interest rate of 5% compounded

continuously. Qasim invested $3,800 in an account paying an interest rate of 6%
compounded annually. To the nearest hundredth of a year, how much longer would
it take for Logan's money to double than for Qasim's money to double?

Answer 1

Explanation:

The formula to calculate the doubling time for continuously compounded interest is:

t = ln(2) / (r * ln(1 + (r/n)))

where t is the time in years, r is the annual interest rate as a decimal, and n is the number of times the interest is compounded per year (which is infinity for continuous compounding).

For Logan's investment, we have:

r = 0.05

n = infinity

t1 = ln(2) / (0.05 * ln(1 + (0.05/infinity)))

t1 ≈ 13.86 years

For Qasim's investment, we have:

r = 0.06

n = 1

t2 = ln(2) / (0.06 * ln(1 + (0.06/1)))

t2 ≈ 11.55 years

To find the difference in the time it takes for their investments to double, we can subtract t2 from t1:

t1 - t2 ≈ 13.86 - 11.55 ≈ 2.31

So it would take Logan's money approximately 2.31 years longer to double than Qasim's money, to the nearest hundredth of a year.