Mistake in solving $^\infty x=2$ for $x$?

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 :)

• 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. Jul 16, 2017 at 9:35
• $$x^{x^{x^{.^{.^{.}}}}}=2\implies x^2=2$$ Jul 16, 2017 at 9:36
• @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? Jul 16, 2017 at 9:38
• Jul 16, 2017 at 9:41
• 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 $b=x^{x^{x^{.^{.^{.}}}}}$
– user436658
Jul 16, 2017 at 9:42

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 $x=1$.

• What do you mean by "the end of $x^{x^{x^{.^{.^.}}}},$ exactly? Jul 16, 2017 at 15:34
• @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. Jul 16, 2017 at 15:45
• I see. I thought you meant the end of the infinitely-long chain of powers of $x$. :-) Jul 16, 2017 at 15:53