I was thinking about the problem $a^{a^{a^{\cdots}}}=2$, I thought of taking $x = a^{a^{a^\cdots}}$, then we get $a^x = x$ but we know as per the question that $x=2$ so we have to solve for $a^x=x$ or $a^2=2$ which we get $a=\sqrt{2}$, but this is also true for $x=4$ for which $a=\sqrt{2}$ also, which seems kind of incorrect?
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How is true for $4$? – Badr B Jun 06 '18 at 03:50
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As we get $a^4=4$ – BAYMAX Jun 06 '18 at 03:52
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@ZacharySelk I see that it contradicts the method but why ? – BAYMAX Jun 06 '18 at 03:54
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1Related: https://math.stackexchange.com/questions/1089458/how-can-i-prove-the-convergence-of-a-power-tower – JMoravitz Jun 06 '18 at 03:56
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1The difference is $x = 2$ is an attractive fixed point for the iteration $x \mapsto \sqrt{2}^x$ while $x = 4$ isn't. If you start from any $x \in [0,4)$, the iterated sequence will converges to $2$. if you start from $x \in (4,\infty)$, the sequence diverges to $\infty$. – achille hui Jun 06 '18 at 04:06
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1https://www.youtube.com/watch?v=DmP3sFIZ0XE&t=295s – mathematics2x2life Jun 06 '18 at 04:49