Solution to $7^{2x-2} \equiv 4 \mod13$ I need to solve the congruence

$$ 7^{2x-2} \equiv 4 \mod13 $$

I think I need to use a primitive root to transform it into $2x-2 \equiv ??? \mod (\phi(13)=12)$ but I'm stumped on how to actually do it.
$$7^{2x-2}\equiv 4 \pmod{13}$$
$\iff$
$$49^{x}\equiv 4(49) \pmod{13}$$
$\iff$
$$10^{x}\equiv 4(-3) \pmod{13}$$
$\iff$
$$10^{x}\equiv 1 \pmod{13}$$



You can now check the powers of ${10} \pmod{13}$.

Alternatively, for a slight shortcut:

Integers mod ${13}$ form a group under multiplication of order $12$. So now you need to find the order of $10$, and you can do this by checking through the factors of $12$.

Then $x\equiv0\pmod{\textrm{order}(10)}$.
$\mod 13$:

$$7^2 \equiv 10 \equiv -3 \\
7^4 \equiv (-3)^2 \equiv 9 \equiv -4\\
 7^8 \equiv (-4)^2 \equiv 16 \equiv 3\\
7^{10} \equiv 7^2\cdot 7^8 \equiv (-3)\cdot 3\equiv -9 \equiv 4
$$
So 
$$
2x-2 = 10 \\
x = 6
$$