I constructed the table for 5 modular 23. While solving the congruence I stuck at point where I have to find Modular inverse of $10\pmod{22}$ which is invalid. I don't know hot to continue from here. Here's how I solved it:
$$7x^{10} + 1\equiv 0\pmod{23}\\ 7x^{10} \equiv 22\pmod{23}$$
applying log and using the table I constructed, I was able to do:
$$\log_5{7} + 10\log_{5}x \equiv\log_{5}22\pmod{22}\\ 19 + 10\log_{5}x \equiv 11\pmod{22}\\ 10\log_{5}x \equiv 14\pmod{22}$$
I can't solve it further because I can't find Modular inverse for $ 10\pmod{22}$. Did I use the wrong method?