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  1. #1
    Groinlonger
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    Fenrir

    Solutions for exponential nonlinear system

    It's been a while since I've done math, perhaps someone can point me in the right direction. I'm trying to figure out ways to solve the following two systems.
    Code:
    e0 = e^(Ax + BxX0) + e^(Ay + ByY0)
    e1 = e^(Ax + BxX1) + e^(Ay + ByY1)
    e2 = e^(Ax + BxX2) + e^(Ay + ByY2)
    e3 = e^(Ax + BxX3) + e^(Ay + ByY3)
    
    e0 = e^(Ax + BxX0) + e^(Ay + ByY0) + e^(Az + BzZ0)
    e1 = e^(Ax + BxX1) + e^(Ay + ByY1) + e^(Az + BzZ1)
    e2 = e^(Ax + BxX2) + e^(Ay + ByY2) + e^(Az + BzZ2)
    e3 = e^(Ax + BxX3) + e^(Ay + ByY3) + e^(Az + BzZ3)
    e4 = e^(Ax + BxX4) + e^(Ay + ByY4) + e^(Az + BzZ4)
    e5 = e^(Ax + BxX5) + e^(Ay + ByY5) + e^(Az + BzZ5)
    e6 = e^(Ax + BxX6) + e^(Ay + ByY6) + e^(Az + BzZ6)
    They are two different systems (not related), but in each system e, X, Y, and Z are known.

  2. #2
    Black Guy from Predator.
    Uppity Negro
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    The Immortal Bill Duke

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    posting for a very particular reason. If you can't figure it out, fuck you.

  3. #3
    The Shitlord
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    Rivendare

    is it because you can no longer see the cinnamon sugar swirls in every bite?

  4. #4
    Groinlonger
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    Fenrir

    I have a theory on what it is.

    In regards to the questions, they can both be solved using numerical approaches. I haven't been able to think of any other way yet.

  5. #5
    Renegade Philosopher
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    What are you trying to solve for?

  6. #6
    BG Content
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    Lakshmi
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    He wants to solve for Ax, Ay, Az, Bx, By, and Bz.

    I took the nearest sharpie to the nearest paper and could only solve for two of the variables for the first equation system in a page.

    Code:
    (ln(e-e^(Ay + By*Y2)) - ln((e-e^(Ay+By*Y1))/X1))*X1/(X1-1) = Ax
    Good luck solving the next two substitutions. Also, knowing how long it has been since I last did math, there's a high likelihood that I already made at least one algebra mistake.

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