Conjecture: $2^{\frac{1}{\log_2(x)}}$ is algebraic iff $x=2^n$ or $1/2^n$ for some $n\in \Bbb Q^+.$
How can I prove my conjecture?
It might be very easy to prove but I am stuck at the moment.
If for example $x=6$ then I think the expression fails to be algebraic due to the fact that $6$ cannot be expressed the above forms.
$3$ is algebraic. And if $x$ is of the form prescribed by your conjecture, then we either have $\log_2(x) = n$ or $\log_2(x) = -n$ for some $n\in \Bbb Q^+$. Regardless of which of them it is, your conjecture would imply that there is a rational $m$ such that $2^m = 3$, which goes against the fundamental theorem of arithmetic.