X^2*log(2)x=256^2 (2) refers to base 2Is ther any rule to solve

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Discussion Overview

The discussion revolves around the equation x^2*log(2)x=256^2, where (2) indicates base 2. Participants explore methods for solving this equation, particularly focusing on the use of the Lambert W function.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that there is no general rule for obtaining an exact solution to the equation.
  • Others propose that the equation can be solved using the Lambert W function, detailing the steps involved in the process.
  • A participant emphasizes that the Lambert W function is not an elementary function but can represent solutions for certain equations.
  • Another participant argues against the dismissal of special functions, suggesting that they should be treated equally in discussions about solving equations.

Areas of Agreement / Disagreement

There is no consensus on the approach to solving the equation. Some participants support the use of the Lambert W function, while others question the necessity or relevance of special functions in this context.

Contextual Notes

Participants express differing views on the classification of functions and their applicability to solving equations, highlighting the complexity of the topic without resolving the underlying disagreements.

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x^2*log(2)x=256^2

(2) refers to base 2
Is ther any rule to solve for x.
 
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No, there isn't any general rule to get an exact solution to that equation.
 


x can be solved for here, but it involves a function that you've probably not heard of, called the "Product Log function", or the "Lambert-W function". It's defined as the inverse function of the function y=x*e^(x). In other words, if x=y*e^(y), then y=W(x).

This is the only way to solve for x in your function, by using Lambert's W Function.

EDIT: Or, if c=1, then x=0.
 


kind said:
x^2*log(2)x=256^2

(2) refers to base 2
Is ther any rule to solve for x.

Here's how to do it via the Lambert W function:

[tex]\frac{x^2}{\log(2)}\log(x)=(256)^2[/tex]

[tex]\log(x)=\frac{\log(2)(256)^2}{x^2}[/tex]

[tex]x=\exp\left\{\frac{\log(2)(256)^2}{x^2}\right\}[/tex]

Now square both sides and divide by x^2:

[tex]1=\frac{1}{x^2} \exp\left\{\frac{2\log(2)(256)^2}{x^2}\right\}[/tex]

[tex]2\log(2)(256)^2=\frac{2\log(2)(256)^2}{x^2}\exp\left\{\frac{2\log(2)(256)^2}{x^2}\right\}[/tex]

and that's where we take the W-function of both sides:

[tex]\frac{2\log(2)(256)^2}{x^2}=W\left[2\log(2)(256)^2\right][/tex]

[tex]x=\pm \sqrt{\frac{2\log(2)(256)^2}{W\left[2\log(2)(256)^2\right]}}[/tex]

Nothin' wrong with the W function is it guys?
 
Last edited:


It is not an elementary function.

To just about any equation you might define a function by which the solution can be represented.

So what?

That Lambert functions, Hankel functions, Bessel functions and whatnot are useful is not the issue here.
 


arildno said:
It is not an elementary function.

To just about any equation you might define a function by which the solution can be represented.

So what?

That Lambert functions, Hankel functions, Bessel functions and whatnot are useful is not the issue here.

Come on Arildno. You made a boo-boo. Just fess-up. He asked how to solve it without qualification and what I showed is conceptually no different than using sines and cosines to solve an equation.

Stop special function discrimination: Equal rights for special functions.
 


jackmell said:
Stop special function discrimination: Equal rights for special functions.


Nope. :smile:
 

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