Question 1202023: how long do you need to invest your money in an account earning an annual interest rate of 2.83% compounded daily so that your investment doubles over that period of time
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! annual interest rate is 2.83%.
if it is compounded daily, then your time period is in days, not years, and your growth factor per year becomes growth factor per day.
assuming 365 days in a year, and assuming a growth factor of (1 + .0283/365), the formula of f = p * (1 + r) ^ N becomes:
2 = 1 * (1 + .0283/365) ^ n.
f is the future value.
p is the presnt value.
r is the growth rate per time period.
1 + r is the growth factor per time period.
n is the number of time periods.
simplify to get:
2 = (1 + .0283/365) ^ n
take the log of both sides of the equation to get:
log(2) = log((1 + .0283/365) ^ n)
by log rule that says log(x^n) = n * log(x), the equation becomes:
log(2) = n * log(1 + .0283/365).
divide both sides of the equation by log(1 + .0283/365) to get:
log(2) / log(1 + .0283/365) = n
solve for n to get:
n = 8940.230697 days.
at 365 days in a year, that becomes 24.49378273 years.
round to two decimal places to get 24.49 years.
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