As stated above: $x^x = \frac{1}{256}$, solve for $x$.
Since $256$ is power of $2$, I let $x = 2^n$, where $n \in R$.
So:
- $2^{n^{(2^n)}} = 2^{-8}$
- $n*2^n = -8$
- $-n = 2^{3-n}$
- $\log_2$$(-n) = 3-n$
- $\log_2$$((-n)^{\frac{1}{3-n}}) =1$
- $(-n)^{\frac{1}{3-n}}= 2$
I am currently stuck at this point with no idea on how I should progress forward. My last option is to solve it numerically but I would like to see if there are any other algebraic ways(other than Lambert W function).
Just to condense comments into an answer:
We seek real solutions for $x$. Let $f(x) = x^x$. Note that $f$ is defined on $\{x > 0\} \cup \{x \in \mathbb{Z}: x < 0\}$.
For $x>0$, by differentiation we can show $f\vert_{x>0}$ has a global minimum at $x = \frac 1 e > \frac{1}{256}$. So $f(x) = \frac{1}{256}$ has no solutions for $x>0$.
Now consider $x \in \mathbb{Z}_{<0}$. Note that $f$ is strictly increasing on this domain, so there is at most one solution. Beginning at $-1$ and descending in steps of $1$, we can find the solution $x=-4$.
(We could have "spotted" $-4$ as well, but this method generalises to other equations $f(x) = y$ where $y \in (0, \frac 1 e]$, either providing a unique solution or the absence of a solution.)