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Math Help - intriguing question..involving logs and exponential decay/growth?

  1. #1
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    intriguing question..involving logs and exponential decay/growth?

    Carbon taken from an old animal skeleton contains 3/4 as much radioactive carbon14 as carbon taken from a present day bone. How old is the animal skeleton? (the half-life of carbon 14 is 5760 years)

    halflife: the time required for something to fall to half its initial value

    let P represent the amount of radioactive carbon 14 from a present day bone

    i have 2 problems with this question.

    i know 1/2P= P(x)^5760, but what adjustment will be made to 5760 since we are dealing with 3/4P instead of 1/2P?

    and how do you isolate x after that?


    thanks in advance
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  2. #2
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    Quote Originally Posted by BuffaloSoulja View Post
    Carbon taken from an old animal skeleton contains 3/4 as much radioactive carbon14 as carbon taken from a present day bone. How old is the animal skeleton? (the half-life of carbon 14 is 5760 years)

    halflife: the time required for something to fall to half its initial value

    let P represent the amount of radioactive carbon 14 from a present day bone

    i have 2 problems with this question.

    i know [1] 1/2P= P(x)^5760, but what adjustment will be made to 5760 since we are dealing with 3/4P instead of 1/2P?

    and how do you isolate x after that?


    thanks in advance
    1. Calculate the base from [1]:

    \dfrac12 P = P \cdot x^{5760} Divide by P and calculate the 5760th root:

    x = \sqrt[5760]{\dfrac12} = \left(\dfrac12\right)^{\frac1{5760}} = 2^{-\frac1{5760}}

    2. According to the question you know:

    \dfrac34 P = P \cdot \left(2^{-\frac1{5760}}  \right)^t

    Divide by P and solve the equation for t. Use the base-change-formula to calculate a logarithm:

    t = \log_{\left(2^{-\frac1{5760}}\right)}\left(\dfrac34  \right)

    Spoiler:
    I've got t \approx 2390\ years
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