Question: I have the following equations :

$$x^5=2 \pmod {13}$$ $$x^3=2 \pmod{11}$$

I wonder how to solve such equations is there a method to get rid off the powers in order to use the Chinese Remainder Theorem without checking each $x$ value?

Any ideas?

Thank you.

Answer: 
${\rm mod}\ 11\!:\ x^{\large 3\cdot 7} \equiv x\,\ $ by $\,\ x^{\large 1+2\cdot 10}\!\equiv x(x^{\large 10})^{\large 2}\!\equiv x(1)^{\large 2}\equiv x\ $ by Fermat (clear if $\,x\equiv 0)$

therefore $\ \ x^{\large 3}\equiv a\iff x\equiv a^{\large 7}\ $

because $\ \left[x^{\large 3} \equiv a\right]^{\large 7}\! \Longrightarrow\, x\equiv a^{\large 7}\ \ \,$ by $\,\ \ x^{\large 3\cdot 7}\equiv x$

because $\ \ x^{\large 3} \equiv a \ \Longleftarrow \ \left[ x\equiv a^{\large 7}\right]^{\large 3} $ by $\,\ a^{\large 7\cdot 3}\equiv a$

**Remark** $\ $ This is essentially the same way your would solve the equation in $\,\Bbb R\,$ by raising $\,x^{\large 3}\,$ to the power $\, 1/3,\,$ but here $\,1/3\equiv 7\pmod{\!10},\,$ and powers can be considered mod $10$ by Fermat. Said functionally, $\,f(x) \equiv x^{\large 3}\,$ is injective $(1$-to-$1),\,$ with inverse $\,x^{\large 7},\,$ because the exponent $3$ is invertible mod $\,p\!-\!1 = 10,\,$ being coprime to it.
we have $$1^5\equiv 1 \mod 13$$
$$2^5\equiv 6 \mod 13$$
$$3^5\equiv 9 \mod 13$$
$$4^5\equiv 10 \mod 13$$
$$5^5 \equiv 5 \mod 13$$
$$6^5 \equiv 2 \mod 13$$
can you proceed?