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This is probably going to be a simple answer, but, how would you convert $x^y-y^x=1$ into $y=$ form without any $y$ on the opposite side of the equation?

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  • $\begingroup$ Not simple at all. There may be some neat little trick to pull it out, but it is definitely not obvious. In general it is easy to write out equations such as this for which is impossible to solve neatly for one of the variables. It can't even be done for polynomials of degree greater than 4. $\endgroup$ – Paul Sinclair Aug 19 '15 at 2:41
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    $\begingroup$ If you graph the solution set in the $x$-$y$ plane (like this) you can see that for some values $y=1.9$ (for instance), there exist multiple $x$ values making the equation true. You can verify this by looking at the intercept of $x^{1.9}-1.9^x-1$ on this plot. This shows that the equation can't be converted to the form $y = f(x)$ -- because then a choice of $x$ value would uniquely determine the $y$ value. $\endgroup$ – vociferous_rutabaga Aug 19 '15 at 2:46
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$x^y-y^x=0$ has a general solution in terms of the Lambert W function. $x^y-y^x=a\neq0,$ however, has no such general solution. If you wish, you may write ${\color{red}y}=\sqrt[\Large x]{x^{\color{red}y}-a}$ , and then repeatedly iterate the expression with regard to y.

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  • $\begingroup$ $y={\displaystyle\root \displaystyle x \of {x^{\displaystyle\root \displaystyle x \of {x^{\displaystyle\root \displaystyle x \of {x^{\displaystyle\root \displaystyle x \of {x^{\displaystyle\root \displaystyle x \of {x^{\displaystyle\root \displaystyle x \of {x^{\cdots}-a}}-a}}-a}}-a}}-a}}-a}}$. (Sorry; I just had to see how TeX would typeset that.) Extra comment: For which values of $x$ does this limit converge? $\endgroup$ – Christopher Carl Heckman Aug 19 '15 at 6:32

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