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Homework Help: Linearization of a non linear equation

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm sure this is easy but I've been looking at it for an hour and can't get anywhere. I have an equation that I need a linear form of.

    2. Relevant equations

    y = a*b*(x1*x22-(x3*x44/c))/(1+b*x1)

    That's the equation I have to write a linear form of.

    3. The attempt at a solution

    I'm struggling to make a start, first thought was to try multiply the LHS by the 1+a*x1 that the RHS is divided by. Then take the natural log of both sides.

    Assuming basic log laws I ended up separating it into:

    2*ln(y) + ln(b) + ln(x1) = ln(a) + ln(b) + ln(x1) + 2*ln(x2) - ln(a) + ln(b) + ln(c) + ln(x3) + 4*ln(x4)

    I know this is wrong since it cancels to

    2*ln(y) = ln (b) + 2*ln(x2) + ln(c) + ln(x3) + 4*ln(x4)

    And that would mean the constant a vanishes as does x1. I guess I'm approaching the linearization wrong, that or my interpretation of log laws is flawed?
    Last edited: Feb 25, 2009
  2. jcsd
  3. Feb 24, 2009 #2

    Tom Mattson

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    FYI this forum is LaTeX-enabled. Just wrap the code in tex tags, or itex tags for inline typesetting. The tags look like the following, only without spaces.

    [ tex ] code goes here [ \tex ]

    [ itex ] code goes here [ \itex ]

    To see the code for any LaTeX image, just click on the image (make sure pop ups are allowed).

    That's not right. For instance, [itex]\ln(y(1+bx_1))=ln(y)+ln(1+bx_1)[/itex]. But you can't separate [itex]ln(1+bx_1)[/itex] any further.

    That said, I'm a bit confused by what you mean by "linearizing" the equation. When I hear that expression I think of finding a linear approximation of [itex]y=f\left(x_1,x_2,x_3,x_4\right)[/itex]. But this necessarily involves calculus, and you placed this in the Precalc forum. Can you explain further what you mean by linearizing an equation?
    Last edited by a moderator: May 4, 2017
  4. Feb 24, 2009 #3
    Ok, thanks for your prompt reply, I'd better explain the context of the problem since I guess I'm using the wrong term but I'm fairly certain that the solution isn't calculus based.

    I've realised my attempt at taking the natural log of each side was wrong, I forgot my basic log laws :blushing:

    That equation is supposed to be rearranged into a form where the constants a and b can be estimated using a function like Microsoft Excels 'LINEST' or similar to do regression analysis. So I thought this meant some sort of rearrangement so it could be approximated to a form such as: [itex]y=mx+c[/itex] or [itex]1/y=m/x+1/c[/itex]. My initial thought was to try and linearize it by the logs. Now I'm thinking it's probably alegraic rearrangement (without logarithms) but I'm not getting anywhere.
  5. Feb 24, 2009 #4

    Tom Mattson

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    Sorry, I don't know how LINEST works. :redface: Also I'm at a loss as to how you could estimate [itex]a[/itex] and [itex]b[/itex] separately, given that they occur in the combination [itex]ab[/itex].

    If you want to use natural logs then I would try the following.


    Now define new variables [itex]u_1=x_1x_2-x_3x_4^4/c[/itex] and [itex]u_2=1+bx_1[/itex].
  6. Jun 1, 2010 #5
    See page 4 in <http://en.wikipedia.org/wiki/Gradient_descent>. [Broken] I think by linearization the poster might mean something like estimating a in y = e**ax using least squares. Take the log base e of both sides, then subject it to linear least squares.
    Last edited by a moderator: May 4, 2017
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