Question 1096195
the formula to use with discrete compounding is f = p * (1+r)^n


when p = 2, the future value will be 1 when the population is halved.


you get 1 = 2 * (1+r)^n


r is the rate of growth per time period.
n is the number of time periods.


in this problem, our time period is days.


divide both sides of the equation by 2 to get 1/2 = (1+r)^n


when n = 11, this formula becomes 1/2 = (1+r)^11


take the 11th root of both sides of the equation to get (1/2)^(1/11) = 1+r


subtract 1 from both sides of the equation to get (1/2)^(1/11) - 1 = r


that's your interest rate per time period.


(1/2)^(1/11) - 1 is equal to -.0610690893


that's the growth rate per day.


test this out with your problem.


when p = 1175, the formula becomes f = 1175 * (1 - .0610690893) ^ 11.


this results in f = 587.5


587.5 * 2 = 1175, therefore 587.5 is exactly 1/2 of 1175.


if you want to know how much of the bacteria is left after d days, then replace 11 by d in the formula to get:


f = 1175 * (1 - .0610690893) ^ d


when d = 11, it's half.


when d = 22, it's half again.


when d = 33, it's half again.


1175 * (1 - .0610690893) ^ 11 = 587.5


1175 * (1 - .0610690893) ^ 22 = 293.75


1175 * (1 - .0610690893) ^ 33 = 146.875


587.5 * 2 = 1175


293.75 * 2 = 587.5 * 2 = 1175


146.875 * 2 = 293.75 * 2 = 587.5 * 2 = 1175


when d = 11, the result is 1/2 of the original.
when d = 22, the result 1/4 of the original.
when d = 33, the result is 1/8 of the original.


the formula you are looking at is:


f = p * (1 - .0610690893) ^ d