Question 1138416: Find the half-life of a radioactive element, which decays according to the function A(t)= A 0A0e - 0.0291 te−0.0291t, where t is the time in years
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula for exponential growth or decay is f = p * e^(rt).
f is the future value
p is the present value
e is the scientific constant of 2.718281828.....
r is the rate per time period.
t is the number of time periods.
i can't figure out your formula very well, so it's hard to fit it into this formula.
you've got A(t)= A 0A0e - 0.0291 te−0.0291t, which i can translate somewhat into:
A(t) = (?????) * e^(-.0291 * t).
assuming you want to find the half life, you can make f = 1/2 and p = 1, so the formula becomes:
1/2 = e^(-.0291 * t)
take the natural log of both sides of this equation to get ln(1/2) = ln(e^(-.0291 * t)).
since ln(e^(-0.0291 * t) is equal to -.0291 * t * ln(e) and since ln(e) = 1, the formula becomes:
ln(1/2) = -.0291 * t
divide both sides of this equation by -.0291 and you get ln(1/2) / -.0291) = t
this results in t = 23.81949074.
the half life of this radioactive element should be equal to 23.81949074 years.
consider the original number being 2 and the future number being half that = 1.
the formula becomes 1 = 2 * e^(-.0291 * 23.81949074 years.) which becomes 1 = 1, confirming the solution is correct.
if your rate per year is .0291, whatever your present value is should be halved in 23.81949074 years.
this is because e^(-.0291 * 23.81949074 = .5, so any value of f = p * e^(-.0291 * 23.81949074) becomes f = p * .5 which means the value of p is halved when the formula is applied.
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