Shank's Baby-Step Giant-Step for $3^x \equiv 2 \pmod{29}$ Problem: Solve $3^x \equiv 2 \pmod{29}$ using Shank's Baby-Step Giant-Step method.
I choose $k=6$ and calculated $3^i \pmod {29}$ for $i=1,2,...,6$.
$$3^1 \equiv 3 \pmod {29}$$
$$3^2 \equiv 9 \pmod {29}$$
$$3^3 \equiv 27 \pmod {29}$$
$$3^4 \equiv 23 \pmod {29}$$
$$3^5 \equiv 11 \pmod {29}$$
$$3^6 \equiv 4 \pmod {29}$$
Then I have calculated $3^{-1} \equiv 10 \pmod {29}$ and started calculating second list:
$$2 \cdot 3^{-6} \equiv 2 \cdot 10^6 \equiv 2 \cdot 22 = 44 \equiv 15 \pmod {29}$$
$$2 \cdot 3^{-12} \equiv 2 \cdot {3^{-6}}^2 \equiv 2 \cdot 15^2 \equiv 2 \cdot 22 = 44 \equiv 15 \pmod {29}$$
$$2 \cdot 3^{-18} \equiv 2 \cdot {3^{-12}}^2 \equiv 2 \cdot 22^2 \equiv 2 \cdot 20 = 40 \equiv 11\pmod {29}$$
And now I can stop. I can see that:
$$ 3^5 \equiv 11 \pmod {29}\ and\ 2 \cdot 3^{-18} \equiv 11 \pmod {29}$$
Therefore $x = 5 + 18 = 23$
But when I plugin $x=23$ above I get that $3^{23} \equiv 8 \pmod {29}$.
So where am I wrong?

I found flaw in my calculus for the second expression in second list, it should be
$$ 2 \cdot 3^{-12} = 2 \cdot (3^{-6})^2 \equiv 2 \cdot 22^2 = 2 \cdot 484 \equiv 11 \pmod {29}$$
Hence, $x = 5 + 12 = 17 $ which is correct.
