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.

$\mod 13$:
$$7^2 \equiv 10 \equiv -3 \\
7^4 \equiv (-3)^2 \equiv 9 \equiv -4\\
 7^8 \equiv (-4)^2 \equiv 16 \equiv 3\\
7^{10} \equiv 7^2\cdot 7^8 \equiv (-3)\cdot 3\equiv -9 \equiv 4
$$
So 
$$
2x-2 = 10 \\
x = 6
$$
