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Math Help - Carban decay.. :(

  1. #1
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    Carban decay.. :(

    i know that the formula is A(t)= Ao-e^-kt ...
    but how do i do this:?
    the half life of carbon-14 is 5750 years. a scroll is found that has lost 22.3% of its carbon-14. How old is the scroll...?
    Thanx
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ijleyton View Post
    i know that the formula is A(t)= Ao-e^-kt ...
    but how do i do this:?
    the half life of carbon-14 is 5750 years. a scroll is found that has lost 22.3% of its carbon-14. How old is the scroll...?
    Thanx
    if the half life is 5750, this means that k = \frac {\ln 2}{5750}

    so your equation becomes: A(t) = A_0e^{- ( \ln 2)t/5750}

    now, we ar econcerned with the time when there is 77.7% remaining, thus we need to solve 0.777A_0 = A_oe^{-( \ln 2)t/5750} for t

    i leave the rest to you
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  3. #3
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    uhm, i dont know if i did this right but the 'e' and 'ln' cancel right?
    and should i get 2233,875?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ijleyton View Post
    uhm, i dont know if i did this right but the 'e' and 'ln' cancel right?
    and should i get 2233,875?
    it is true that e^{\ln x} = x, but that is a totally different situation from what we have here. we have something like e^{-( \ln x)/K}, different things come into play here. that being said, your answer is incorrect. don't try anything fancy, just take the \ln of both sides and work it out using regular algebraic manipulation. because you would need to change the function a bit to use the e^{\ln x} = x rule, and to me, it's not worth it here.
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  5. #5
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    so: k=(about) -1.2054 x 10^-4 and
    .777= e ^kt --> ln both sides
    ln (.777) = ln (e ^kt) --> Today i friend old me you could bring the kt down like:
    ln (.777) = Kt ln(e) --> If thats correct then t = 2093.077..
    is that right :-s?

    Edit: If bringing down the exponent is right can someone explain why you're allowed to do that?
    Last edited by ijleyton; January 10th 2008 at 06:12 PM. Reason: forgot something
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  6. #6
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    so: k=(about) -1.2054 x 10^-4 and
    .777= e ^kt --> ln both sides
    ln (.777) = ln (e ^kt) --> Today i friend old me you could bring the kt down like:
    ln (.777) = Kt ln(e) --> If thats correct then t = 2093.077..
    is that right :-s?

    Edit: If bringing down the exponent is right can someone explain why you're allowed to do that?
    You can indeed bring the exponent down. The definition of a log is:
    if y = x^n then n = \log_x(y)

    y = x^n also implies that y^p = x^{np}, so \log_x(y^p) = np<br />
    but n = \log_x(y) so \log_x(y^p) = p\log_x(y)
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