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I came across the equation $2^{x} = x^{x} + x$ that was claimed to be only solvable via graphing or approximation. Is that the case, and if so why? I don't understand why the answer can't be described with operations.

Arturo Magidin
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Ammar Aldabbagh
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    no, such equations are not solvable algebraically, Although some functions like lambert function are defined, But they are still only named because such equations are non-solvable by simple algebraic techniques, Although, the trivial solution by guess work is $x=1$ – Dheeraj Gujrathi Nov 15 '23 at 08:56
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    $x=1$ is a solution for example. – Sebastiano Nov 15 '23 at 08:57
  • @DheerajGujrathi is there a way to prove that such question cannot be solved algebraically? I'll look up the lambert function. – Ammar Aldabbagh Nov 15 '23 at 08:59
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    @Sebastiano that's guess and check, what if the question was a bit more complicated such that it is not intuitive. Also, the other solutions are not as simple, and I think(?) that -1.30854... is not even rational, so how did we reach that answer. – Ammar Aldabbagh Nov 15 '23 at 09:02
  • @AmmarAldabbagh The equation is no linear. Hence you could use the Lagrange method to find the approximation solutions. – Sebastiano Nov 15 '23 at 09:06
  • Existence of algebraic solution is probably special case, for example for polynomial equations, the Abel-Ruffini theorem says that there is no general algebraic solution for polynomials with order >= 5

    https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

     – lizardfireman Nov 15 '23 at 09:09
  • @Sebastiano from a visit to Wikipedia, it seems that the Lagrange method is a way to guess the answer more precisely (could that be described as an 'algorithm'?). My question is why can't the answer be described in terms of what Wikipedia calls 'elementary functions'. – Ammar Aldabbagh Nov 15 '23 at 09:15
  • Probably, Newton's method can be use here. – Peter Nov 15 '23 at 09:19
  • And since we are in the reals (I assume) , we should also restrict $x$ to be positive. – Peter Nov 15 '23 at 09:21
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    @DheerajGujrathi If you allow special functions, the equation is solvable algebraically. But if you disallow them, why are you not disallowing logarithm functions? Then 2^x=5 would not be solvable. Why are you not disallowing cube roots? Then x^3=2 is not solvable. So where do you draw the line? – Divide1918 Nov 15 '23 at 09:36
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    @AmmarAldabbagh the reason the Lambert W function can't be expressed as an elementary function is non-trivial. You can find more here. https://mathoverflow.net/questions/135911/how-to-prove-lamberts-w-function-is-not-elementary/ – CyclotomicField Nov 15 '23 at 09:42
  • @Divide1918, The problem is we can't draw a line, I understand what you mean to say, But I Assumed OP meant to say was using simple algebraic manipulation generally taught in highschools or even during undergrad courses – Dheeraj Gujrathi Nov 15 '23 at 09:47
  • $2^{x} = x^{x} + 3x$ could be a better example as it does not have an obvious solution (approximately $0.08$) while $2^{x} = x^{x} + x$ does ($1$) – Henry Nov 15 '23 at 10:04
  • As for cube roots, they are just fractional powers, so there is no new function except for new notation and domain extension. Also, is the asker also looking for complex solutions? – Тyma Gaidash Nov 15 '23 at 13:40
  • There are various numerical method techniques to approximate the solution to as much precision as needed... – Nothing special Nov 15 '23 at 18:01
  • @ТymaGaidash I don't have much background with complex numbers but my question is to do with principle, not the answer to this particular problem. If solving with complex numbers would help me, then sure I'm fine with allow that to be the scope. What makes it so it is not solvable, and how do I find out if an equation is or isn't solvable. – Ammar Aldabbagh Nov 19 '23 at 06:28

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