# Thread: How to solve congruence equation with unknown power? [11^a=9 (mod 23)]

1. ## How to solve congruence equation with unknown power? [11^a=9 (mod 23)]

Hi everyone. I am struggling with solving the following equation.

11^a=9 (mod 23)

I know that we have to use logarithms here, but do not know how to implement it... Could someone explain me how to solve it. Thank you in advance.

2. Originally Posted by Denis
Geesshhhh...did you know that 9 mod 23 = 9?

11^a = 9
a = LOG(9) / LOG(11)

Take over, Rover

What is 30 mod 23?

Bro, I know it, This equation has the answer, and it is 6. I want to know how to solve it, not just estimating.

3. Originally Posted by newtagi
11^a=9 (mod 23)

I know that we have to use logarithms here, but do not know how to implement it... Could someone explain me how to solve it. Thank you in advance.
Are you studying something like what they discuss here? Also, are you working with "discrete" logs?

Thank you!

4. Originally Posted by stapel
Are you studying something like what they discuss here? Also, are you working with "discrete" logs?

Thank you!
You are right man) I have searched for it in the internet, and I found couple algorithms that are based on trivial mutiplication. My teacher told me, if I solve this problem, I do not need to participate in final exam. Now I understand why she said like that.

5. Originally Posted by newtagi
You are right man) I have searched for it in the internet, and I found couple algorithms that are based on trivial mutiplication. My teacher told me, if I solve this problem, I do not need to participate in final exam. Now I understand why she said like that.
Here is a discussion of this sort of problem, going into some depth for both a small problem like yours, and a large one (which would easily take the place of a final exam).