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Math Help - find half life, simple problem

  1. #1
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    find half life, simple problem

    A 10 g sample of tritium decays to 8.5 grams in 2.93 years. What is the half life of tritium

    the formula is y= A (0.5)^t/h

    so A is the original amound, t is the time, and h is the half life, y is the changed amount...

    i tried doing it but i didnt know...can anybody help? Does it involve logarithms?

    this is what i got:

    8.5 = 10 (0.5) ^2.93/h

    log 8.5 = log (10/2) ^ 2.93/h

    h = 2.93 log 8.5 / log 10/2
    Last edited by magmagod; April 21st 2010 at 02:10 PM.
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  2. #2
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    Quote Originally Posted by magmagod View Post
    A 10 g sample of tritium decays to 8.5 grams in 2.93 years. What is the half life of tritium

    the formula is y= A (0.5)^t/h

    so A is the original amound, t is the time, and h is the half life, y is the changed amount...

    i tried doing it but i didnt know...can anybody help? Does it involve logarithms?
    Set t = 2.93, y = 8.5, and solve for h. And yes, there are logarithms involved. And of course set A = 10.

    Edit: fixed typo
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  3. #3
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    i know that...i meant for somebody to answer it cause i tried and failed in doing so.. i edited my question..you can check it again if you want
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  4. #4
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    Quote Originally Posted by magmagod View Post
    i know that...i meant for somebody to answer it cause i tried and failed in doing so.. i edited my question..you can check it again if you want
    You are close.

    Here is my work:

    8.5 = 10\left(0.5\right)^{\frac{2.93}{h}}

    0.85 = \left(0.5\right)^{\frac{2.93}{h}}

    log(0.85) = log\left((0.5)^{\frac{2.93}{h}}\right)

    log(0.85) = \frac{2.93}{h}log(0.5)

    \frac{2.93}{h} = \frac{log(0.85)}{log(0.5)}

    h*log(0.85) = 2.93*log(0.5)

    h = \frac{2.93*log(0.5)}{log(0.85)}
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  5. #5
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    ohhh...i get it so you just need to divide by 10 in the begining, then log both sides..thanks
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