# How do I find X of this equation?

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?

-
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. – Shaun Ault Oct 18 '11 at 0:00
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... – J. M. Oct 18 '11 at 0:01
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! – Shaun Ault Oct 18 '11 at 0:01
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 $x$. – Brian M. Scott Oct 18 '11 at 0:03
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. – André Nicolas Oct 18 '11 at 0:05

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.