In that case the solution is unique if 
and it doesn't exist otherwise.
Think about the graph if you want, the function is strictly increasing from 0 to infinity. (I'm assuming p/q>0, you can play around with the other cases)
Of course sometimes you can extend the domain if you want like x^n we define on the whole reals (and complex too if you like, you get n solutions) and then you get other solutions. But in general when defining x^r we like to restrict to the non-neg reals(sometimes even non-negative complex) otherwise we get some issues. That's what I meant by it depending on what planet you live on.
Correct me if I'm wrong guys, but for this question to make any sense for the purposes of VCE maths, I can only assume that p/q is a rational number, therefore
x^(p/q) = a => x = a^(q/p)
ignoring the trivial case p=0, a=1, x in R, we further require q not 0
hence the relevant cases are:
1a) q even, p positive => a in R
1b) q even, p negative => a in R\{0}
2a) q odd, p odd and positive => a in R
2b) q odd, p odd and negative => a in R\{0}
3a) q odd, p even and positive => a in R+ U {0}
3b) q odd, p even and negative => a in R+
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