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Solving logarithmic equations

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data
    0ops.. The title should read solving logs..
    Mod note: Fixed.

    the original dosage contains 280 MBq of Iodine-131. If none is lost from the body, then after 6 hr there are 274 MBq of iodine-131. What is the half life of iodine I-131?

    2. Relevant equations

    Logcx=y

    3. The attempt at a solution

    I tried to put
    274=280(1/2)^t/6 in and solve for t. But now I know it can't be right because it is not halving every 6 hrs.. However I don't know how you would figure it out. Can someone give me a hint?
     
    Last edited by a moderator: Nov 11, 2013
  2. jcsd
  3. Nov 11, 2013 #2
    The concentration varies with time according to the equation [itex]I=I_0e^{-kt}[/itex], where k is a constant that you need to determine from the data: -kt = ln(I/I0). Once k is known, find the time at which I is equal to half of I0: -kt1/2=ln(1/2)
     
  4. Nov 11, 2013 #3

    Ray Vickson

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    Put ##D = D_0 (1/2)^{t/c},## where ##D =## dosage after t hours, ##D_0 = ## initial dosage (at t = 0) and ##c## is a constant. You are given ##D_0 = 280##, and at t = 6 you have ##D = 274##. You have enough information to find ##c##.
     
  5. Nov 11, 2013 #4

    Dick

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    The radioactivity will follow the decay law 280*(1/2)^(t/k). k is the half life. You want to solve for k, not put k=6.
     
  6. Nov 12, 2013 #5
    Thank you!
     
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