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 :)
 A: 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.
A: 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 $x=1$.
