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Consider the hyperpower equation $$x^{x^{x^{x...}}}=2$$ We will use the method that

Let $y=x^{x^{x^{x...}}}$ so $x^y=x^{x^{x^{x...}}}=2$ and $x^2=2$ to give the solution $x=\sqrt2$

However consider another equation such $x^{x^{x^{x...}}}=4$

Use the similar method so we have $y=x^{x^{x^{x...}}}$ to give $x^y=x^4=4$ , this also makes $x=\sqrt2$ as our solution

But the $\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2...}}}$ have to equal to something, not 2 and 4 at the same time.

If there is only one solution $x=\sqrt2$, then what is wrong with the second equation?

Steve
  • 876

1 Answers1

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This is known in mathematics as Tetration. It is proven that the maximum value that $y$ can be is $e$ which implies that $x=e^{1/e}$. For $x>e^{1/e}$, the infinite tetration does not converge. So my point is that your argument of $x^{x^{x^{x^\ldots}}}=4$, does not make any sense.