If y=x^x then what is x in terms of y?

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Discussion Overview

The discussion revolves around the equation y = x^x and the challenge of expressing x in terms of y. Participants explore whether this can be achieved algebraically or through other means, including the use of the Lambert W function.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that finding x in terms of y can be done through iteration, while questioning the possibility of an algebraic solution.
  • One participant proposes taking the logarithm of both sides, leading to the equation ln(y) = xln(x), and introduces the Lambert W function as a potential solution, stating that x = W(ln(y)).
  • Another participant notes that standard functions do not suffice for this problem, hence the relevance of the Lambert W function.
  • A participant poses a related question about the inverse scenario, asking what y would be if x = y^y, and applies the same logarithmic approach to derive y = W(ln(x)).
  • There is a clarification that the previous manipulation simply involves swapping x and y, not altering the fundamental approach.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of an algebraic solution, with some supporting the use of the Lambert W function while others question the algebraic nature of the approach. The discussion remains unresolved regarding the broader implications of these manipulations.

Contextual Notes

The discussion highlights the limitations of standard functions in solving the equation and the specific conditions under which the Lambert W function is applicable. There is also an implicit assumption regarding the definitions and properties of the functions involved.

3hlang
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i know this is possible to find with iteration, but is it possible to find it algebraically?
 
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Hi 3hlang! :smile:
3hlang said:
i know this is possible to find with iteration, but is it possible to find it algebraically?

Nope!
 
This probably isn't what you mean by "algebraically" but if you take the logarithm of both sides of the equation you get ln(y)= xln(x) and then x= W(ln(y)) where W is the "Lambert W function" which is defined as the inverse function to xln(x).
 
... and the reason we talk about the Lambert W function is that none of the previously-defined standard functions would work.
 
so if x=y^y, then y=?
 
3hlang said:
so if x=y^y, then y=?

HallsofIvy said:
This probably isn't what you mean by "algebraically" but if you take the logarithm of both sides of the equation you get ln(y)= xln(x) and then x= W(ln(y)) where W is the "Lambert W function" which is defined as the inverse function to xln(x).

All you have done is swap x and y! If [itex]x= y^y[/itex] then, by the same formula I gave before, y= W(ln(x)).
 
what? i wasn't claiming to have done anything radical to the formula. all i wanted to know was what you do the w function of. but thank you for telling me
 

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