Why do $x^{x^{x^{\dots}}}=2$ and $x^{x^{x^{\dots}}}=4$ have the same positive root $\sqrt 2$? What is  $x$ when is satisfies $x^{x^{x^{\dots}}}=2$ ?


I am really confused with this; the root is $\sqrt{2}$, but why does the equation $x^{x^{x^{\dots}}}=4$ have the same root?
Let's first start with defining what we mean by the expression $a = x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}$: the infinite power tower is the limit of the recursion

$$a_{n+1} = x^{a_n},~~~~~~~ a_0 = x$$

If $x&gt;1$ then $x^x &gt; x$ so $a_{n}$ is monotonically increasing and converges iff it's bounded above (this is the case only if $e^{-e} \leq x \leq e^{\frac{1}{e}}\approx 1.44$; see e.g. Wiki:Tetration).  If $x=\sqrt{2}$ then it follows by induction that $a_n \leq 2$ since $a_0 = \sqrt{2} &lt; 2$ and 

$$a_{n+1} = \sqrt{2}^{a_n} \leq \sqrt{2}^2 = 2$$

Thus $a_n$ is bounded above by $2$ (and the recursion therefore converges). Taking the limit of the recursion we get that it satsify the equation $a=x^a$. As you have noted this equation has the two solutions $a=2$ and $a=4$ (which are also the only real solutions). Since the limit is well defined only one of these two solutions are valid and since $a\leq 2$ we have that the solution $a=4$ does not describe the limit of the recursion above.

Why is there an extra solution? Note that

$$\lim\limits_{n\to\infty} a_n = a \implies a = x^a$$

is a one way implication and it's not true that $a = x^a \implies \lim\limits_{n\to\infty} a_n = a$. This is similar to

$$x=1 \implies x^2 = 1 \implies x = \pm 1$$

In this simple example it's squaring that introduces an extra solution. This happens quite a lot when manipulating equations: if not all of the steps are two-way implications then we might introduce extra solutions and we must use some other means to determine which one is the right one to choose.



In general when the infinite tower converges the limit is given by

$$a = -\frac{W(-\log(x))}{\log(x)}$$

where $W$ is the (principal branch of the) Lambert function. The maximum value (for $x$ where it converges) is found for $x = e^{\frac{1}{e}}$ where we have $a = e &lt; 4$. Thus $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} = 4$ has no solutions at all in real numbers so the manipulations you do to get $x = \sqrt{2}$ are not justified.
Assuming the limit exists, then since $$x^{x^{x^{\cdots }}} = 2$$ then it follows that $$x^2 = 2$$ which has two solutions: $x = \sqrt{2} $ and $x = -\sqrt{2}$. If you're working over the reals, then you'll want to ignore $-\sqrt{2}$. 

Do note that all this operates under the assumption that the limit exists, you need to prove it does before the working above is valid.