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Math Help - Radioactive Decay

  1. #1
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    Radioactive Decay

    A radioactive material is known to decay at a yearly rate proportional to the amount at each moment. There were 2000 grams of the material 10 years ago. There are 1990 grams right now. What is the half-life of the material?

    The equation to solve is P = P0e^kt. Where P = population, P0 = initial population, k is a constant and t = time.

    To find k:
    1990 = p0e^0k
    -------------------
    2000 = p0e^-10k

    Once I divide these out I get 1990/2000 = e^10k.
    To find k take the natural log of both sides:-

    ln 1990/2000 = 10k. Therefore k = 1/10*ln 1990/2000.

    To solve for the half life:

    1000 = 2000e^(1/10 ln 1990/2000)t.

    When I work through this I get:-
    t = 10* (ln 1/2)/(ln (1990/2000)).

    However the correct answer is:-
    t = 10* (ln 2)/(ln (2000/1990)).

    Much appreciated if someone can show where I am going wrong.
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  2. #2
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    Quote Originally Posted by p75213 View Post
    When I work through this I get:-
    t = 10* (ln 1/2)/(ln (1990/2000)).

    However the correct answer is:-
    t = 10* (ln 2)/(ln (2000/1990)).
    Both answers equal to the same number
    Just use the fact that ln(1/x)=-ln(x) and you can see that your number is equal to
    t = 10* (-ln 2)/(-ln (2000/1990)) = 10* (ln 2)/(ln (2000/1990)).
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  3. #3
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    Thanks for pointing that out.
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