Primitive roots used to work out $x^7 \equiv 5 \pmod {11}$ I have a workbook question that doesn't have any example solution, that is as follows:
Primitive roots used to work out $x^7 \equiv 5 \pmod {11}$
Now I can see $\phi(11)=10$ and $2$ has order $10$ so $2$ is the primitive root $\pmod{11}$
I can see aswell that $2^4 \equiv 5 \pmod{11}$, so I can show $x^7 = 2^4 \pmod{11}$, must I compute the powers of all the numbers until I find which yields $5 \pmod{11}$, how do I use primitive roots to solve the prolem?

Let $x=2^k$. Then solve for $k$.
ADD We have $2^{7k}=2^4\mod 11$, so that $2^{7k-4}=0\mod 11$. Since the order of $2$ modulo $11$ is $10$, $7k=4\mod 10$. Hence $k=3\cdot 4=12=2\mod 10$. It follows $x=4$.
