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Logarithmic Half-Life prob.

  1. May 6, 2009 #1
    1. The problem statement, all variables and given/known data
    The half-life of a radioactive substance is 194 days. How many days will it take for 80% of the substance to decay?


    2. Relevant equations



    3. The attempt at a solution
    I had to make 1/2n=1/5. I did this by finding log25=n. After this, I needed to only multiply by the number of days it took the half-life to occur, which was 194, and got 451 (approx.) as a reasonable answer.
     
    Last edited: May 6, 2009
  2. jcsd
  3. May 6, 2009 #2

    diazona

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    There is absolutely enough information there. I could tell you the answer right now (but I won't, that's what you get to do ;-)

    I'd advise you to go back and double-check that equation, though.
     
  4. May 6, 2009 #3

    HallsofIvy

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    In one "half life", T, (1/2)C= CekT. The Cs cancel, and you solve that for k. After you know that you can solve 0.8C= Cekt for t.

    But you don't need to use "e". If T is the half life then [itex]C(t)= C(0)(1/2)^{t/T}[/itex] because every time "T" days, you multiply by1/2. Solve [itex].8C= C(1/2)^{t/194}[/itex].
     
    Last edited: May 6, 2009
  5. May 6, 2009 #4

    diazona

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    Looks good...
     
  6. May 6, 2009 #5
    =) omg! Thanks!
     
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