Question 1187575
the radioactive form of uranium has a half life of 2.5 * 10^5 years.


the formula is f = p * e^(rt).


r is the rate of growth per year.
t is the time in years.


if the future value of the uranium is equal to half the present value of the uranium, then the formula becomes:


1/2 = e^(rt)


if the time to half life is equal to 2.5 * 10^5 years, then the formula becomes:


1/2 = e^(r * 2.5 * 10^5)


take the natural log of both sides of the equation to get:


ln(1/2) = ln(e ^ (r * 2.5 * 10^5))


since ln(e ^ (r * 2.5 * 10^5)) is equal to r * 2.5 * 10^5 * ln(e) and since ln(e) = 1, the formula becomes:


ln(1/2) = r * 2.5 * 10^5


divide both sides of the equation by (2.5 * 10^5) to get:


ln(1/2) / (2.5 * 10^5) = r


solve for r to get:


r = -2.77258872 * 10^-6


to see if that value is good, replace r in the original equation with that and find f.


the original equation is f = p * e^(rt)


when p = 1, the equation becomes:


f = e^(rt)


when r = -2.77258872 * 10^-6 and t = 2.5 * 10^5, the formula becomes:


f = e ^ (-2.77258872 * 10^-6 * 2.5 * 10^5)


solve for f to get:


f = .5.


the formula is good and the rate of growth is -2.77258872 * 10^-6 per year.


the remaining mass of 1 gram of uranium after t years is given by the formula:


f = 1 * e ^ ((-2.77258872 * 10^-6 * t)


when t = 5000, the formula becomes:


f = 1 * e ^ ((-2.77258872 * 10^-6 * 5000)


solve for f to get:


f = .9862327045 grams.


you can graph the equation.


here's what it looks like.
it shows the values of f (represented by y in the graph) for t (represented by x in the graph) at 5000 years and at 2.5 * 10^5 years in the future.


<img src = "http://theo.x10hosting.com/2021/111111.jpg" >