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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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    Why did the term "( 5 * X )" become "( 5X * X^2 )" in the second line? If you are multiplying both sides of the equation by X, then that term should only become "( 5X * X )". A similar remark applies to the last term.2011-10-18
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    Due to the logarithm, you have a transcendental equation. Due to the variable being both inside and outside the logarithm, deriving the explicit solution will be difficult. Due to the dissimilarity of $5x+\frac{10}{x}$ and $x$, I doubt that there's a solution in terms of the Lambert function...2011-10-18
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    Moreover, you seem to have "multiplied by $X$", and then "divided by $X$" (regardless of the algebra mistakes). Thus it's no coincidence that you get back the same equation -- okay, well it actually is a huge surprise, given that you made a combination of algebra mistakes that nevertheless undid one another!2011-10-18
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    Your multiplication by $x$ is incorrect: $x(5x) = 5x^2$, not $5x^3$, and $x(10\log x) = 10x\log x$, not $10x^2\log x$. You actually multiplied the third and fourth terms on the righthand side by $x^2$, not by $x$. Your new equation should have been $10x=4x+10+4x^2+10x\log x$, at which point you could have subtracted $4x$ from both sides to get $6x=10+4x^2+10x\log x$. This is not a nice equation, however, and you’ll not be able to solve it for $x$.2011-10-18
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    The multiplication by $X$ was not done right. You should get $10X=5X+10+5X^2 +10X\log(X)$. And there is no nice "formula" for solving this equation.2011-10-18
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    Note that when trying to solve an equation, it’s never useful to perform an operation and then perform exactly the inverse operation, as you’ve done here in first multiplying and then dividing by $x$: if the operations are done correctly, you’re **guaranteed** to be going ’round in circles. It’s exactly like adding and then subtracting $1$: there’s no point.2011-10-18
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    @BrainMScott, yes I see the error of my ways there. I just was getting frustrated at not being able to figure this out. Why am I not able to go backwards? If I know every outcome going forward why can't I backtrack?2011-10-18
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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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    Hmmm I see. What do you mean 'find a solution numerically'?2011-10-18
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    "Find a solution numerically" means use a systematic approach to find a solution to two decimal places, or five decimal places, or whatever accuracy you want/need. One very good way to do this is to use Newton's Method, which requires Calculus, but another way is "educated trial and error", where you try some values, work out roughly where a solution is, then try more and more values to pin it down more and more accurately.2011-10-18
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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