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Find D

  1. Nov 6, 2015 #1
    How do I get D by itself? This one's got me baffled

    [(A - D)^P] / [(B - D)^R] = (C - D)^(P - R)
     
  2. jcsd
  3. Nov 6, 2015 #2

    SteamKing

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    Take a lotta logs and see if anything shakes out.
     
  4. Nov 7, 2015 #3
    Ha ha ha! I've been to the bathroom enough times while trying to figure this out.
     
  5. Nov 7, 2015 #4
    I have been trying to come up with a definition for Pascal's Triangle so that I can create a general way to solve for unknown exponents. All that I've been able to come up with so far is (1 - N + (N^2 - N)/2 - [1/2∑(n=2 to N) N(N - 2n + 1) + n(n - 1)] . . . ) for Pascal's Triangle of Coeficients
     
  6. Nov 7, 2015 #5

    SteamKing

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    Unknown exponents of what?
     
  7. Nov 7, 2015 #6
    My exponents, P and R
     
  8. Nov 7, 2015 #7

    fresh_42

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    Basically you have x^r * y^(p-r) = 1. Without knowing anything about r and p it'll going to be hard. Are you dealing with economic indexes?
     
  9. Nov 7, 2015 #8
    I don't know what economic indexes are, but I am trying to solve for two unknowns in the common equation: V(t) = Vf + (Vi - Vf)*e^(-t/T) where Vf and T are unknown. I am trying to fit this curve to data points that I have collected.
     
  10. Nov 7, 2015 #9

    SteamKing

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    That's a completely different equation than what you had in the OP.
     
  11. Nov 7, 2015 #10
    I want to fit the curve to three points, one point gives my Vi, the other two are my different V(t)s. I simplified to get: T = -t/[ln((V(t) - Vf)/(Hi - Hf))]. Now I can set two equations equal to each other: -ta/[ln((V(ta) - Vf)/(Hi - Hf))] = -tb/[ln((V(tb) - Vf)/(Hi - Hf))].
    This simplified to [(H(ta) - Hf)/(Hi - Hf)]^tb = [(H(tb) - Hf)/(Hi - Hf)]^ta. Then [(H(ta) - Hf)^tb]/[(H(tb) - Hf)^ta] = (Hi - Hf)^(tb-ta)
     
  12. Nov 7, 2015 #11

    WWGD

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    It would help if you could Tex this, making it easier to read.
     
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