Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
Hmmm I see. What do you mean 'find a solution numerically'? –  AedonEtLIRA Oct 18 '11 at 0:10
1  
"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. –  Gerry Myerson Oct 18 '11 at 0:19
    
I see, thank you. –  AedonEtLIRA Oct 18 '11 at 0:20
    
You might look at chapter 9 of apps.nrbook.com/c/index.html or under "one dimensional root finding" in any numerical analysis textbook. –  Ross Millikan Oct 18 '11 at 0:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.