Solving $7^x\bmod {29} = 23 $ I have $$7^x\bmod {29} = 23 $$
It is possible to get $x$ by trying out different numbers but that will not be possible if $x$ is actually big.
Are there any other solutions for this equation?
Kind regards

I don't think there is anything but trial and error.
But look.  $7^2 \equiv 23 \equiv -6 \pmod {29}$
$7^{2x} \equiv 36 \equiv 7\pmod {29}$.
So $7^{2x-1} \equiv 1 \pmod {29}$.
We know by FLT the $7^{28}\equiv 1$ so for the least power to which $7$ is a multiple of $1$ must be a divisor of $28$.
So $2x-1$ is a multiple of a divisor of $28$.  Immediately we have $2x-1$ is odd so it must be a multiple of an odd divisor of $28$.
The only nontrivial odd divisor of $28$ is $7$.  So test to see if $7^7\equiv 1 \pmod {29}$.  We have $7^7 \equiv 1\pmod {29}$. That was lucky. so we have $2x-1 = 7k$ and  as there are only $7$ values of $7^w\pmod {29}$, we might as well assume $x \le 6$. so $x = 4$ is only option.  Let's try that.
$7^4 \equiv 23\pmod {29}$.
Yep.... that worked.
