Find the value of $x$ if $x^{x^4} =4$ Find the value of $x$ if $x^{x^4}=4$. Given options are 


*

*$2^{1/2}$

*$-2^{1/2} $

*Both 1. & 2

*None of the above


From option verification, we get option 3. as correct one. But is there any real method to do the above problem?
 A: There is no systematic way of solving such transcendental equations. In this particular case, we can get better insight by performing a transformation to get rid of the double exponentiation.
$$4=x^{x^4}=(x^4)^{x^4/4}=t^{t/4}.$$
Then
$$t^t=4^4$$ and an obvious solution is $t=4$ corresponding to $x=\pm\sqrt2$.
By the study of the function $t^t$, we can verify that it is increasing where it exceeds $1$, so that the above solution is unique.
A: $x\in\mathbb{R}$
$(x^{-4})^{x^4}=4^{-4}$ => $x^{-4}=4^{-1}$ => $x^4=4$ => $x=\pm\sqrt{2}$ 
EDIT: 
Jyrki Lahtonen gave me the advise to carry out the proof accurately.
Therefore: $x^{x^4}=4$ , $(z;a):=(x^{-4};\frac{1}{4})$ => $z^{\frac{1}{z}}=a^{\frac{1}{a}}$  
We have $0<a^{\frac{1}{a}}<1$ therefore we have only one positive solution $z=a$. 
This means $x^4=4$ and therefore $x=\pm\sqrt{2}$.
EDIT 2:
$x^\frac{1}{x}$ is strictly increasing (therefore bijective) for $0<x<e$ because of 
$(x^\frac{1}{x})’=x^{(-2+\frac{1}{x})}(1-\ln x)>0$. 
A: You already know how to verify the options here, but you're asking for a way to solve this equation step-by-step. The issue is that there is no way to find a closed form solution for $x$ using elementary functions. The solution is only expressible using a special function known as the Lambert W. This function is the inverse of the function $f(x) = xe^x$. You can read more about it here.
There are two ways to handle this "from first principles".
The first, which has already been covered by @user90369, is to find solutions by observation (which means inspired guess and check, really), then prove rigorously that no other solutions can exist.
The second is to find a direct solution using the Lambert W.
We start by simplifying the form of the equation with the substitution $x = y^{\frac 14}$
After some elementary simplification, you will end up with $y^y= 256$. This is, of course, very amenable to a solution "by inspection", but let's proceed as originally intended.
$$y^y = 256$$
$$e^{\ln y e^{\ln y}} = 256$$
$$\ln y e^{\ln y} = \ln 256$$
$$\ln y = W(\ln 256)$$
$$y = e^{W(\ln 256)} = \frac{\ln 256}{W(\ln 256)}$$
where a property of the Lambert W is used for the final step.
At this point, you have to use a special calculator or mathematical software to find out the value of $W(\ln 256)$. One such calculator is here.
Using that, we can do the calculation to find that $y$ is very close to $4$. We can now make an "inspired guess" that it is $4$ since no more exact calculation is available to us. We find that it works, so we accept the solution (you should know how to find $x$ after determining $y$).
A: I'm not sure if this is rigorous, but you could notice that because $4=x^{x^4}$, we can write 
$$4=x^{x^4}=x^{x^{x^{x^4}}}$$
and so on, until we get a stack of $x$'s equalling to $4$:
$$x^{x^{x^{x^{.^{.^{.}}}}}}=4$$
And so 
$$x^4=4$$ 
which yields $x=\pm\sqrt 2$
A: $x^{x^4}=4$
First of all, we think simple.
$x^{x^4}=2^{2^1}$
We did that because we have to make first two steps equal$x$
Later we have to make our third step in our second equation ,4 .
$$x^{x^4}=2^(2^(1/4))^4$$
$$x^x=2^{2^{1/4}}$$
We have simplified our problem. Now we should make both steps equal in our second equation. If we take the square of our power number (to make sure they are equal), we must take the square root of our number. Therefore:
$2^{2^{1/4}}= 2^{{1/2}^{2^{1/2}}}$
Our previous power $4$ was even. 
Solutions are $-2^{1/2}$ and $2^{1/2}$.
