Could we find an element on finite field?

Question

Let $F$ be a finite field. Given an element $a^x$ in $F\setminus\{0\}$, could we find $a$?? I know that finding an integer $x$ is very hard problem (Discrete Logarithm Problem). However, I don't know the hardness of this problem. What's the problem's name? Thanks.

Answer 1

$0^x$ is in the field for every non-zero $x$.

Answer 2

Let $q = |F|$. Since $(F^*,\times)$ is a cyclic group of order $q-1$, you are basically asking when and how we can divide by $x$ in $\Bbb Z/(q-1)\Bbb Z$.

This depends heavily on $x$. If $x$ and $q-1$ are not coprime, then this is impossible (for example if $x=q-1$ well $a^x= 1$ forall $a \in F^*$ so knowing $a^x$ gives you absolutely no information on $a$).

If they are, then $x$ has an inverse $y$ modulo $q-1$, and dividing by $x$ is the same as multiplying by $y$ : in the context of $(F^*,\times)$ this means that $a = (a^x)^y$.