$$x^{x^{x^{.^{.^{.}}}}} = 8$$

Then how to solve for x?

I first tried like this

$x^8=8$ but I don't get any way to solve.

  • 1
    $\begingroup$ I don't see why solving $x^8=8$ does not work. It has $2$ real solutions, namely $2^{3/8}$ and $-2^{3/8}$. $\endgroup$ – Levent Dec 23 '16 at 17:06
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    $\begingroup$ Well, you also have to show that $\large{x_{n+1} = x^{x_n}}$ converges. $\endgroup$ – GFauxPas Dec 23 '16 at 17:07
  • $\begingroup$ @Levent do we need to take log? $\endgroup$ – Fawad Dec 23 '16 at 17:09
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    $\begingroup$ Related: 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. $\endgroup$ – Winther Dec 23 '16 at 17:10

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.

  • $\begingroup$ Can you explain a bit in detail how you got that inequality? $\endgroup$ – Fawad Dec 24 '16 at 2:59

Notice that

$$x^8=8\\\implies x=\sqrt[8]8\approx1.29683955465$$

And since


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.

  • $\begingroup$ Can I know how you got that inequality? $\endgroup$ – Fawad Dec 24 '16 at 3:00
  • $\begingroup$ Did you try reading the link in my answer? $\endgroup$ – Simply Beautiful Art Dec 24 '16 at 12:45

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