<s>[INST] Solving a congruence using primitive roots Suppose we know that $3$ is a primitive root of $17$. 
How can that help us solving $7^x \equiv 6 \pmod {17}$?
 [/INST] 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, $3^{(17-1)/2}\equiv-1\pmod{17}$
$\implies11x\equiv14+1\pmod{16}\equiv-1$
$\implies3\cdot11x\equiv-1\cdot3\pmod{16}$
$\iff x\equiv-3\equiv13\pmod{16}$
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