Why does x raised to an odd power result in a unique solution? I don't think I'm phrasing this correctly but anyway, say we are looking for solutions to x for $x^9=1/2$,
why is there only one unique solution as given by $x=(1/2)^{1/9}$
when by raising x to an even value gives two solutions as $\pm$ solution
I think I'm just missing a key concept so any help would be appreciated.

$x^9=1/2$ has exactly 9 solutions by the Fundamental Theorem of Algebra, however the principal root is $2^{1/9}$
Raising $x$ to the even power, say $x^2=1$ would make it so there can be two solutions $\pm1$, since $x^2-1=0$ evaluates to $(x+1)(x-1)=0$, so there are positive AND negative roots. Another way to think about this is if you plug $-1$ into $x^2=1$ for $x$, you have $(-1)*(-1)=1$, which is correct.
On the other hand, if you raise $x$ to an odd power, say $x^3=1$, you obtain only 1 real solution, which is $1$. $-1$ does not work because $(-1)*(-1)*(-1)=-1$, not $1.$
