I recently came across the problem: $$x^{x^{x^{.^{.^.}}}}={^\infty x}=2$$ My approach was to take logs of both sides: $$\ln{2}=\ln{^\infty x}$$ and therefore: $$\begin{eqnarray*}\ln{2}&=&x\ln{^\infty x}\\ & = & x\ln{2}\end{eqnarray*}$$ which would imply that $x=1$. Clearly this is not true, and I have seen many other ways of doing it which produce the correct answer of $\sqrt2$.

Is my mistake that you can't use logarithms as normal when dealing with infinitely stacked powers? If not where else have I gone wrong?

Thanks for any help :)

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    $\begingroup$ The solution to this equation can be defined only when $x^{x^{...}}$ is at least a number, since you are taking the logarithm of this quantity, and furthermore using a logarithmic law as well on it. Hence, it wouldn't be correct to use it unless you state where it converges. Once it does, all laws become applicable. $\endgroup$ Jul 16, 2017 at 9:35
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    $\begingroup$ $$x^{x^{x^{.^{.^{.}}}}}=2\implies x^2=2$$ $\endgroup$ Jul 16, 2017 at 9:36
  • $\begingroup$ @астонвіллаолофмэллбэрг since we have said it is all $=2$ does this mean that we can say it converges and as such can use logarithmic laws? If so I must have gone wrong somewhere else in my working? $\endgroup$ Jul 16, 2017 at 9:38
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    $\begingroup$ youtube.com/watch?v=JrOG1tKAatg. $\endgroup$ Jul 16, 2017 at 9:41
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    $\begingroup$ Your mistake is to use a logarithmic "law" that doesn't exist in nature: $\ln a^b$ is not $a\ln b,$ it's $b\ln a$ (with $a=x$ and $b=x^{x^{x^{.^{.^{.}}}}}$ $\endgroup$
    – user436658
    Jul 16, 2017 at 9:42

2 Answers 2


Parentheses can be tricky

It is important how you set the parentheses! I think you mean: $$x^{\Big( x^{\big(x^{({{{\dots}}})}\big)} \Big)} = 2.$$ So according to these parentheses you get $$\ln\left(\color{red}{x}^{\color{blue}{\Big( x^{\big(x^{({{{\dots}}})}\big)} \Big)}}\right) = \ln(2 )$$ Now by applying the rule $\ln(\color{red}{a}^{\color{blue}{b}}) = \color{blue}b\ln(\color{red}{a})$ we conclude $$ \color{blue}{x^{\Big( x^{\big(x^{({{{\dots}}})}\big)} \Big)}} \ln(\color{red}{x}) = \ln(2) $$ after substitution with you initial assumption you obtain $$2 \ln(x) = \ln(2)$$ This leads to $x=\sqrt{2}$ after some easy manipulations.

More formal Level

To take this to a more formal level you are regarding the recursive sequence $$x_{n+1} = x^{x_n} \quad\text{with}\quad x_1 = x \quad\text{for some}\quad x \in \mathbb{R}^{+}$$ However, if you set the parentheses like $$\bigg(\Big(\big(x^x\big)^x\Big)^{x}\bigg)^{\dots}$$ then you have another recursive sequence: $$x_{n+1} = (x_{n})^x \quad\text{with}\quad x_1 = x \quad\text{for some}\quad x \in \mathbb{R}^{+}$$ If you want to solve $\lim_{n\to\infty} x_n = 2$ for the second recursion, you will come to $$2 = \lim_{n\to\infty} x_n = \lim_{n\to\infty} x_{n+1} = \lim_{n\to\infty} (x_n)^x = \big(\lim_{n\to\infty} x_n \big)^x = 2^x$$ and then you conclude $x=1$ but only if there is a solution for this problem you know that it must be $1$. So this doesn't mean that there is a solution!

There are three cases for your starting point $x$

  • $x\in (0,1)$: In this case $x_n$ is monoton increasing and the limit is $1$.
  • $x \in \{1\}$ In this case $x_n$ is constant and the limit is $1$.
  • $x \in (1,+\infty)$ In this case $x_n$ is monoton increasing and unbounded so we say the limit is $+\infty$.

Now as you can see the sequence can either have $0,1$ or $+\infty$ as a limit.


Instead of taking $\log$, you can think in this way:

What you are doing is $x^{x^{x^{.^{.^.}}}}=2\Rightarrow 2^x=2\Rightarrow x=1$

Similarly you could do this:

$x^{x^{x^{.^{.^.}}}}=2\Rightarrow 2^{2^x}=2\Rightarrow 2^x=1\Rightarrow x=0$

This implies $1=0$, which is absurd. So the process you are using is clearly wrong.

The proper way is:

$x^{x^{x^{.^{.^.}}}}=2\Rightarrow x^2=2\Rightarrow x=\sqrt{2}$ (since positive)

You can think it as a sequence of number reaching the number $2$.

Let $a_1=x,a_2=x^x,a_3=x^{x^x},\dots$. So $a_n=x^{x^{^{.^{.^{.^x}}}}}$($n$-times)

Then the problem is here $\lim_{n\to\infty}a_n=2$, then find $x$.

What is wrong in your process:

a^{b^c}$\neq$ (a^b)^c. Equality only holds when $c=1$ As you can see if you start from the end of $x^{x^{x^{.^{.^.}}}}$, you are doing a^{b^c}$=$ (a^b)^c, which gives $x=1$.

  • $\begingroup$ What do you mean by "the end of $x^{x^{x^{.^{.^.}}}},$ exactly? $\endgroup$ Jul 16, 2017 at 15:34
  • $\begingroup$ @CameronBuie by "the end of ..." I mean way procedure OP used, Sorry, I am not good at english. More precisely I mean if OP write $2^x=2$, then he implies a^{b^c}$=$(a^b)^c. $\endgroup$
    – MAN-MADE
    Jul 16, 2017 at 15:45
  • $\begingroup$ I see. I thought you meant the end of the infinitely-long chain of powers of $x$. :-) $\endgroup$ Jul 16, 2017 at 15:53

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