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

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The equation y = x^x can be transformed using logarithms, leading to ln(y) = xln(x). This results in x being expressed in terms of the Lambert W function as x = W(ln(y)). Conversely, if x = y^y, then y can similarly be expressed as y = W(ln(x)). The Lambert W function is necessary here because standard functions cannot solve the equation algebraically. The discussion emphasizes the iterative nature of finding solutions while clarifying the algebraic manipulation involved.
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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 x= y^y 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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