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Math Help - Logarithmic Functions, Radioactive Decay.

  1. #1
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    Nebraksa
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    Logarithmic Functions, Radioactive Decay.

    The half-life of radium is 1,690 years. how long will it take for a 50-gram sample of radium to be reduced to 5 grams? The answer should be 5,614 years.

    The formula I'm using is Q(t)=Q0e-kt and I believe I am to use h= ln2/k. Here's what I've done:

    5=50e(-k)(1,690)
    .1=e(-k)(1,690)
    ln .1=(-k)(1,690)
    I divide by 1,690 and get -5.917...

    I'm not sure where to use the h= ln2/k, I tried putting it in for k, it didn't look right so I erased it all and went back to square one. Any help is greatly appreciated. :]
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  2. #2
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    Re: Logarithmic Functions, Radioactive Decay.

    You don't say what it is, but I think that h is the half life, 1690. In that case, the formula h = ln2/k allows you to calculate the value of k.
    However, why bother to remember one more formula when you don't have to. Better, (from the point of view of not cluttering your memory with formulae you don't need), is to work from the basic formula Q(t)=Q_{0}(t)e^{-kt}. That way, you get a better understanding of what is going on.

    You're told that the half life is 1690 years, so 25=50e^{-1690k}.
    (This is the equation that gives rise to 'the other' equation h = ln2/k).
    Solve it for k.

    When there are 5 gram remaining, 5 = 50e^{-kt}.
    Solve this for t, substituting for k on the way.
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  3. #3
    MHF Contributor

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    Re: Logarithmic Functions, Radioactive Decay.

    Given that the half life is 1690, you can simply write that as S(t)= S(0)(1/2)^{t/1690}. That is, t/1690 tells "how many" half lifes and then we multiply the original amount by 1/2 that many times. So you want to solve 5= 50(1/2)^{t/1690}. Divide both sides by 50 to get (1/2)^{t/1690}= 1/10 and take the logarithm of both sides.
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