Linearization of a non linear equation

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?
 
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Tom Mattson

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EDIT: realised that might not be the clearest form to interpret, this is the equation i was trying to write above:
http://g.imagehost.org/0914/equation1.png [Broken]
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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)
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?
 
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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?
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.
 

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.

[tex]\ln(y)=\ln(ab)+\ln\left(x_1x_2^2-\frac{x_3x_4^4}{c}\right)-\ln\left(1+bx_1\right)[/itex]

Now define new variables [itex]u_1=x_1x_2-x_3x_4^4/c[/itex] and [itex]u_2=1+bx_1[/itex].
 
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.
 
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