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This question is a branch of my previous question.

I'm trying to reverse an equation. I did everything I thought I was suppose to do, but I reached an impass. I have no idea how to reduce passed initial / X an get to removing the antilog (that too I'm not sure how to remove).

I started with (the values are faked):

10 = 5 + ( 10 / X ) + ( 5 * X ) + ( 10 * log( X ) ) 

I then tried to remove X from the denominator.

10X = 5X + 10 +( 5X * X^2 ) + ( 10X * log( X )X ) 

Then I divided the X from the multiplication parenthesis.

10 = 5 + ( 10 / X ) + ( 5 * X )  + ( 10 * log( X ) ) 

If I did everything correctly, I haven't done anything to this equation. All I can figure out to do is just recurssively add and remove * X to each of these terms. Further once I finally break it down I'm not sure how to remove the log( X ), but that is another question I think.

What am I missing to cancel out the X's?

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    @AedonEtLIRA: It's not that you can't backtrack: it's that *it's a waste of time*. If you do something, and then *undo* it, then you didn't do anything overall. Why bother doing it and then undoing it, then? You only introduce the possibility of making an error along the way. Why take three steps forward and then three steps back, making it possible for you to trip and fall, if you can just stay in place instead?2011-10-18

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Going from the first equation to the second, when you multiply $(5*X)$ by $X$, you should get $(5*X^2)$ instead of $(5X*X^2)$ and when multiplying $(10*\log(X))$ by $X$ you should get $(10X*\log(X))$. Going from the second to the third you reverse the errors.

That said, when you have polynomial terms and logarithmic terms in the same equation, as here, you generally cannot find an algebraic solution unless you like the Lambert W function. You can find a solution numerically.

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    You might look at chapter 9 of http://apps.nrbook.com/c/index.html or under "one dimensional root finding" in any numerical analysis textbook.2011-10-18