How can I solve $x^{5}\equiv 13 \pmod{18}$. I need to solve the congruence of $ x^{5}\equiv 13 \pmod{18}$ and find whether it is solvable.

Actually I can solve $x^{k}\equiv \pmod{q}$ when $q$ is odd. But now it is composite, and I do not know how to apply the theorem. Can you help me?
Hint: By Euler–Fermat, $x^{6}\equiv 1 \bmod{18}$.
By Euler's theorem, we get $13x\cong1\pmod{18}$.  So invert $13\pmod{18}$.  $7\cdot13-5\cdot18=1$ (*the extended euclidean algorithm can be used for this).  Thus $13^{-1}\cong7\pmod{18}$.  Thus we get $x\cong7\pmod{18}$.

*$\begin{align}18=13+5\\
13=2\cdot5+3\\
5=3+2\\
3=2+1\end{align}$

So,  $1=3-2=3-(5-3)=2\cdot3-5=2(13-2\cdot5)-5=2(13)-5(5)=2(13)-5(18-13)=7(13)-5(18)$.