# 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 $x=4$.
• So I have $(2^k)^7=2^{7k} = 2^4$ then using a theorem result, $7k=4 \pmod{10}$, $7$ has inverse $3$, so $k = 7*3*4$, so $x^7 = x^{84}$? - – Display Name May 4 '14 at 21:27
• Edit was too late above: " So I have $(2^k)^7=2^{7k} = 2^4$ then using a theorem result, $7k=4 \pmod{10}$, $7$ has inverse $3$, so $k = 7*3*4$, so $2^7 \cong 5 \mod{11}, 2^{84} = x^7, x = 2^{12}$?" – Display Name May 4 '14 at 21:35
• Could you please tell me why $x$ must be of that form and not anything else? – rah4927 May 5 '14 at 4:54