Can Mathematica turn (y+x)=x^(1/y) into y=f(x)? [on hold]

I wanted to use Mathematica to turn a few implicit equations into explicit expressions. This worked fine for most of my equations, yielding mostly simple results and a few with a Lambert w function (aka product log) thrown in, but for some reason it won't solve

(y + x) == x^(1 / y)


to give me $y=f(x)$. Now I'm not the best versed person in Mathematica, so it may not be huge when I say I've tried everything I can think of, but I'm definitely stuck and out of Ideas. Can anybody help?

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It doesn't look like it's possible. Reduce[y + x == x^(1/y), y] simply gave a generic warning that it's not solvable with the methods available to Solve or Reduce. –  DumpsterDoofus Feb 8 at 17:52
I don't think there's a closed form solution in terms of "simple" functions. In these cases it's good to think about whether you really do need such a closed form solution. What would it enable you to do? You can already calculate the solution numerically. You might define your own $f(x)$ as a solution of this equation and try to prove identities for it which may be useful for your work. Having a closed form solution in itself has no value unless it's written in terms of functions that you can look up and find many useful identities for. –  Szabolcs Feb 8 at 19:34
Raise both sides to the y power and you get (x + y)^y == x. Consider a case like y=2... this is a second order polynomial and has two answers for y=2 -- hence a form like f[y] is impossible because it is not a single-valued function. –  bill s Feb 9 at 4:05
bill - you have a point but explicit functions aren't necessarily single valued. Many are multivalued such as product logs. –  Mike Feb 9 at 19:37
Szabolcs - an explicit closed-form function would make things incredibly easier and may be necessary in the future, but I may sate it as an evaluation of newton's method if I have to. –  Mike Feb 9 at 19:40