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paulr said:
You can solve that equation approximately, e.g. if y=100 then x=3.5972850236. See a book about numerical analysis to learn how to do that. Basically given y, you take a guess at x, then if x^x is too big, make x smaller; and if x^x is too small, make x bigger. Using a computer, keep improving the guess til it's accurate enough for your purpose.
You could also write out an infinite series solution, but it would be messy. There will not be any simple formula like y=sqrt(sin(x)) or anything like that.
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Exactly. I have done some more looking around on this. Several authors cite this exact example as a problem that cannot be expressed that way.
One guy suggested making a new function for this exact dilemma. Then you could answer this in simple algebraic terms.
When you think of it, it's not that far out. We currently have cube roots/etc that are very useful functions (on a calculator/etc) that are very hard to do by hand. If this was a more common problem, I am sure my HP calculator would have a button for it.
Just for the record, to anyone who said "square root": that's not right.