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Re-arranging equation: negative time for exponential

  1. Apr 20, 2013 #1
    1. The problem statement, all variables and given/known data
    I have a sequence whereby
    10000=100(1+e^(kt)+e^(2kt)+...+e^(39kt)) where k=-4.7947012×10^(-3) which was dervied from dy/dt=ky
    Re-arranging i get 99=1+e^(kt)+e^(2kt)+...+e^(39kt), letting e^(kt)=r I put it into the computer and
    i get 1.04216=r=e^-4.7947012×10^(-3)t
    taking ln of both sides and dividing by the number gives a negative time value. Any help is highly appreciate.
     
  2. jcsd
  3. Apr 20, 2013 #2

    mfb

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    With a negative k, t has to be negative, too. That is a correct solution of the equation. If a negative time value is impossible, your equation (or k) has to be wrong.

    There is an analytic solution, by the way.
     
  4. Apr 20, 2013 #3
    Analytical, how so?
     
  5. Apr 20, 2013 #4
    Is it by stating that, if assuming that all doses haven't decayed, the maxmimum amount is 3900 only.
     
  6. Apr 20, 2013 #5

    mfb

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    With r=e^(kt), your equation gets 99=r^0 + r^1 + r^2 + ... + r^39
    That is a geometric expression, and has a nice formula.
     
  7. Apr 20, 2013 #6

    Ray Vickson

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    Right. And that gives you a degree = 40 polynomial to solve---not easy at all, and I doubt there is an analytical formula for its solution.
     
  8. Apr 20, 2013 #7

    mfb

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    Hmm, you are right. Well, at least it is easier to solve it numerically that way (which does not matter if a computer solves it).
     
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