Question 1054018
the continuoous compounding formula is:


f = p * e^(r*n)


f is the future value
p is the present value
e is the scientific constant of 2.718281828
r is the interest rate per time period.
n is the number of time periods


in your problem:


p = 14018
r = .063 per year
n = t years


the formula becomes f = 14018 * e^(.063 * t)


you want to know what is the doubling time.


if 14018 doubles, then it is worth 28036.


the formula becomes 28036 = 14018 * e^(.063 * t)


divide both sides of the equation by 14018 to get 28036 / 14018 = e^(.063 * t)


simplify to get 2 = e^(.063 * t)


take the natural log of both sides of the equation to get ln(2) = ln(e^(.063 * t).


since ln(e^x) = x*ln(e), your equation becomes ln(2) = .063 * t * ln(e).


since ln(e) = 1, your equation becomes ln(2) = .063 * t


divide both sides of the equation by .063 to get ln(2) / .063 = t.


solve for t to get t = ln(2) / .063 = 11.0023362


thqt's the number of years it would take for the money to double at 6.3% per year using continuous compounding.


14018 * e^(.063 * 11.0023362) is equal to 28036.