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Nice solution Davis. It is also interesting to find maximum x for which the sequence \( x^{x^{x^{...}}} \) converges.

Agree with you. Here is some easier example: 1 2 3 3 1 2 2 3 1

Hi, Davis try to solve \( x^{x^{x^{...}}} = 4 \) by your method. It seems the answer will be the same \( {\sqrt2} \) . Am I correct?

I believe the answer for task 4 is 0.75 and here is the solution uploaded which uses geometric probability:

Bob, are you sure about \(x = \sqrt{2}\) I believe we can use such a logic only if the sequence has a limit. In other case if I use the same logic to equation: \(a=x^{x^{\ldots}}=4\) I will get: \(ln(x^{x^{\ldots}}) = ln4\) \(x^{x^{\ldots}}lnx = ln4\) \(a lnx = ln4\) \(4lnx =...