$$x^{x^{x^{.^{.^{.}}}}} = 8$$
Then how to solve for x?
I first tried like this
$x^8=8$ but I don't get any way to solve.
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1I don't see why solving $x^8=8$ does not work. It has $2$ real solutions, namely $2^{3/8}$ and $-2^{3/8}$. – Levent Dec 23 '16 at 17:06
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1Well, you also have to show that $\large{x_{n+1} = x^{x_n}}$ converges. – GFauxPas Dec 23 '16 at 17:07
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@Levent do we need to take log? – Fawad Dec 23 '16 at 17:09
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1Related: How can I prove the convergence of a power-tower? In summary the power-tower converges for all $x\in[1/e^e, e^{1/e}]$. $2^{3/8}$ is inside this interval. – Winther Dec 23 '16 at 17:10
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Notice that
$$x^8=8\\\implies x=\sqrt[8]8\approx1.29683955465$$
And since
$$e^{-e}<x<e^{e^{-1}}$$
then it converges to the proposed number.
It then remains that none of the other solutions to $x^8=8$ are possible, which is explained in this answer.
Simply Beautiful Art
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There is no way to obtain an analytical solution in terms of elementary functions.
However, one can find an expression in terms of the Lambert W function. This expression evaluates to $8=-\frac{W(-ln(x))}{ln(x)}$. This expression can be solved using numerical methods.
However, as noted by others you may notice that infinite tetration (technical word for power-tower) of $x$ converges if and only if $x \in [e^{-e},e^{1/e}]$. Therefore, your may use your positive real solution to $x$ for $x^8=8$. This will be identical to the solution to your question since: $$e^{-e}<x<e^{1/e}$$ Therefore, $x$ cannot converge to any other value.
projectilemotion
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