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Thread: Math B problem (please help)

  1. #1
    Jun 2005

    Math B problem (please help)

    this is a Math B problem for June 2004 # 31. I have the answersheet but i cant figure out how they got the answer.

    31 An archaeologist can determine the approximate age of certain ancient
    specimens by measuring the amount of carbon-14, a radioactive substance,
    contained in the specimen. The formula used to determine the
    age of a specimen is A = A02^(-t/5760) , where A is the amount of carbon-14
    that a specimen contains, A0 is the original amount of carbon-14, t is
    time, in years, and 5760 is the half-life of carbon-14.
    A specimen that originally contained 120 milligrams of carbon-14 now
    contains 100 milligrams of this substance. What is the age of the
    specimen, to the nearest hundred years?
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  2. #2
    MHF Contributor
    Apr 2005
    This is just a substitution problem.

    A = A02^(-t/5760)
    Rewriting that,
    A = (Ao)*[2^(-t/5760)]

    Let us say that is correct.
    (It is. I am more used to A = (Ao)*[(1/2)^(t/5760)] in this case.)

    Ao = 120 mg
    A = 100 mg
    So, just substitute them into the equation.
    Remember, we need to isolate the t, so we do anything to isolate it.

    100 = (120)*[2^(-t/5760)]
    Divide both sides by 120,
    100/120 = 2^(-t/5760)
    0.83333333333 = 2^(-t/5760)
    Get the logarithms (common log is okay) of both sides,
    log(0.8333333333) = (-t/5760)*log(2)
    -0.0791812 = (-t/5760)*(0.30103)
    -0.0791812 = t(-0.30103/5760)
    -0.0791812 = t(-0.00005226)
    t = (-0.0791812)/(-0.00005226)
    t = 1515 years

    Therefore, to the nearest hundred, the specimen is 1,500 years old. ----answer.

    You can use the natural log, or Ln, if you like, and the answer will be the same.

    No need to solve it that long if you can skip steps. If you're good with calculators, you need only one or two steps to get t at once.
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