Prove that if $x^x=4$ then $x=2$. Trying to prove that I looked at $x\ln x=2\ln 2$ and stated that therefore $x=2$... It was just a little assumption in a big question. Now I see that $x^x$ is a composition of injections and therefore is an injection, and since $2^2=4$ then no other $x$ satisfies it. Am I relatively correct? I would appreciate your help. 
 A: The function $x^x$ is undefined for $x\le 0$.
It is easy to show that $x^x\le 1$ for $0<x\le 1$.
The derivative of $x^x$ is $x^x(\ln x+1)$ which is positive for $x>1$.
Combined, these facts say that if $x^x=4$ then $x>1$. The function is increasing in that interval, so there can be at most one solution there. Therefore, there is at most one solution anywhere, and $x=2$ is it.

Here is a method, using the Lambert W function, that can easily be extended to solve the equation $x^x=a$ uniquely for any $a\ge 1$.
$$x^x=4$$
$$x\ln x=\ln 4$$
$$e^{\ln x}\ln x=\ln 4$$
$$\ln x=\operatorname{W}(\ln 4)$$
$$x=e^{\operatorname{W}(\ln 4)}$$
The Lambert W function $\operatorname{W}(x)$ is single-valued for $x>0$, and certainly $\ln 4>0$. So $x$ is well defined and can be calculated by a variety of software. We end up with the value $x=2$ in our particular case. This method shows that the value of $x$ has at most one solution in $x^x=a$ for $a\ge 1$.
(That list of software leaves out the dmath package for Borland/Embarcadero Delphi, as well as the Graph application.)
A: By definition, $z^w = e^{w \ln z}$. Set $e^{w \ln z} = 4$ with the condition that $w = z$ and solve it.
$e^{x \ln x} = 4 \Rightarrow x = \frac{\ln 4}{\ln 2} \Rightarrow x = 2$.
A: Suppose $x\gt y \gt 1$ then $y^y\lt x^y\lt x^x$ so for $x\gt 1$ the function $x^x$ is increasing, and there is at most one solution in this range - the existence of a solution is not in doubt.
If $0\lt x\lt 1$ we have $x^x\lt 1$ and $1^1=1$.
A: Hint: You can show that $$f(x)=\frac{1}{\log_4(x)}$$
has an only fixed point.
