Is it computationally infeasible to solve $x^a\equiv b \pmod p$ as that in discrete logarithm problem to solve $a^x\equiv b \pmod p$, where $p$ is a large prime and $a,b$ are both positive integer below $p1$
If $a$ is relatively prime to $p1$, you can solve for $c$ from $ac\equiv 1$ (mod $p1$) and raise both sides to the power $c$. You obtain $x\equiv x^{ca}\equiv b^c$ (since $x^{p1}\equiv 1$) (mod $p$). If $a$ is not relatively prime to $p1$, the congruence is not always solvable. However, there should be a least $d$ for which $ac\equiv d$ (mod $p1$) has a solution and raise both sides to the power $c$. Then one has the congruence $x^d\equiv b^c$ (mod $p1$). In most cases this will be simpler than the original congruence. The congruence has no solution if $b$ is not a $d$th power modulo $p$. It should be clear that $d$ will be a divisor of $p1$ (since it can be computed as the gcd of $a$ and $p1$) and that $b$ will be a $d$th power if $b^{p1/d}\equiv 1$ (mod $p1$). That is as far as I can help. 


There are probabilistic algorithms to solve your equation with complexity polynomial in $\log p$. That means that you can solve such equations in practice. You can find the algorithms in any book on computational/algorithmic number theory. Funnily enough, there is no known deterministic such algorithm even for $a=2$. 

