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Logarithim question decreasing in value

  1. Dec 2, 2009 #1
    Logarithim question "decreasing in value"

    1. The problem statement, all variables and given/known data

    The Radioactivity (R) of a substance can be modelled using the equation:

    R = A x 2 ^ -Bt

    Where R is measured in becquerels per gram, t is measured in hours and A and B are constants.
    If a substance has an initial radioactivity of 60 becquerels and one hour later it is 50 becquerels find A and B and the time for the radioactivity to reduce to half its initial value (Half Life)



    2. Relevant equations

    R = A x 2 ^ -Bt



    3. The attempt at a solution

    R = A x 2 ^ - Bt

    R = 60 t = 0
    R = 50 t = 1
    60 = A x 2 ^ - Bx0

    60 = A

    50 = 60 x 2 ^ - B

    2 - B = 50/60

    2 - B = 0.8

    log 2 - B = log 0.8

    - B log 2 = log 0.8

    - B = log 0.8/log 2

    - B = - 0.32

    Is this all right?

    I work out half life to be around 3 hours but could anybody give me a more exact time?

    Thanks in advance
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 2, 2009 #2
    Re: Logarithim question "decreasing in value"

    Solved t = 3.125

    Thanks
     
  4. Dec 2, 2009 #3

    Mark44

    Staff: Mentor

    Re: Logarithim question "decreasing in value"

    The next line is incorrect. It should be 2-B = 50/60 = 5/6
    The next line is also incorrect. 5/6 != .8
    The next line is incorrect, but at least I understand what you are doing. It would be correct if you had on the right side (log 5/6)/log 2
    You're actually off by quite a bit. To 4 decimal places, B = .2630
    It's about 3.8 hours.
     
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