Solving a congruence using primitive roots

Question

Suppose we know that $3$ is a primitive root of $17$.

How can that help us solving $7^x \equiv 6 \pmod {17}$?

Answer 1

Use Discrete Logarithm wrt primitive root $3\pmod{17},$

$x$ind$_37\equiv$ind$_3(2\cdot3)\pmod{\phi(17)}\equiv$ind$_32+1$

Now $3^3\equiv10\pmod{17},3^5\equiv9\cdot10\equiv5\implies7\equiv5^{-1}\equiv3^{-5}\equiv3^{16-5}$

$2\equiv(-1)3\cdot5\pmod{17}\equiv3^{8+1+5}$ as $3$ is a primitive root, $3^{(17-1)/2}\equiv-1\pmod{17}$

$\implies11x\equiv14+1\pmod{16}\equiv-1$

$\implies3\cdot11x\equiv-1\cdot3\pmod{16}$

$\iff x\equiv-3\equiv13\pmod{16}$