Find the value of $x$ if $x^{x^4} =4$

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.

$x\in\mathbb{R}$

$(x^{-4})^{x^4}=4^{-4}$ => $x^{-4}=4^{-1}$ => $x^4=4$ => $x=\pm\sqrt{2}$

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 $(x^\frac{1}{x})’=x^{(-2+\frac{1}{x})}(1-\ln x)>0$.

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 $y$).