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Thread: 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 $\displaystyle k = \frac {\ln 2}{5750}$

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

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

    i leave the rest to you
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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 $\displaystyle e^{\ln x} = x$, but that is a totally different situation from what we have here. we have something like $\displaystyle e^{-( \ln x)/K}$, different things come into play here. that being said, your answer is incorrect. don't try anything fancy, just take the $\displaystyle \ln$ of both sides and work it out using regular algebraic manipulation. because you would need to change the function a bit to use the $\displaystyle e^{\ln x} = x$ 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; Jan 10th 2008 at 05:12 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 $\displaystyle y = x^n$ then $\displaystyle n = \log_x(y)$

    $\displaystyle y = x^n$ also implies that $\displaystyle y^p = x^{np}$, so $\displaystyle \log_x(y^p) = np
    $
    but $\displaystyle n = \log_x(y)$ so $\displaystyle \log_x(y^p) = p\log_x(y)$
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