Solve $7^x \equiv 6 \pmod{17}$ given 3 is a primitive root $\bmod 17$ It's easy to show that 3 is a primitive root $\bmod 17$,
but how do I use it prove the congruence? 
Is there a general way to solve any congruence of the form $a^x \equiv b \pmod{c}$ if you know a primitive root $\bmod c$ and c is big (without brute force)?
 A: Here if you observe that $7=3^{11}\quad \& \quad 6=3^{15}$ so Putting above we have
$3^{11x}\equiv 3^{15}~ \pmod{17}$
$\implies 3^{11x-15}\equiv 1 \pmod{17}$
3 is primitive root so  $11x-15=16n$ for some $n\in \mathbb{Z}$
$\implies x=\dfrac{16n+15}{11}$ will be integer
You can easily see that $n=8$ satifies this ,So $x=13$ will be a solution of given .
There will be more solution of this too.
A: You can reduce the amount of brute force quite significantly. Using an interval $d$ of approximately $\sqrt{c}$, make an "island" of values $a\cdot g^i$ with $i\in \{0,d{-}1\}$, then calculate all $g^{jd}$ and identify the coinciding value.
$\begin{array}{c|c}
k\ (d=4) &g^k & a\cdot g^k & b\cdot g^k \\ \hline
0  & 1 & 7 & 6 \\
1  & 3 & \color{red}{4} & \color{violet}{1} \\
2  & 9 & 12 & 3 \\
3  & 10 & 2 & 9 \\
4  & 13\\
8  & 16\\
12  & \color{red}{4}\\
16  & \color{violet}{1}\\
\end{array}$
giving, $\bmod 17$, $7\cdot 3\equiv 3^{12}$ and $6\cdot 3\equiv 3^{16}$ , so $7\equiv 3^{11}$ and  $6\equiv 3^{15}$. 
Then we need to find $11x\equiv 15 \bmod 16 (=\phi(17))$, which means finding the inverse of $11\bmod 16$, which here is $3$, giving $x\equiv 15\cdot 3 \equiv 13\bmod 16$.
Obviously for this exercise in small numbers this is more work thn simply calculating the powers of $7 \bmod 17$, but for larger numbers it avoids calculating all powers.
Because $3$ is a primitive root $\bmod 17$, we are guaranteed to find suitable exponents in the first stage.
