Ah, it looks like, as with that issue, you can get solve to give you
an answer if you pass check=False:
In [21]: print(solve(-(S(21)/20)**(12*x) + 120, x, check=False))
[-log(120**(1/(-log(7355827511386641) + log(4096000000000000))))]
which looks like the same answer after some log identities are applied
(those large numbers are 21**12 and 20**12, respectively).
Aaron Meurer