0 does work as a trivial solution, since 0^a = 0^b always holds, but yea, 30's the only other one - the key comes down to the fact that a^log(b) = b^log(a) in general, and 30 is what achieves that here
In your own work, you divided by x^ln(6) at one point, which would remove any x = 0 solution from your work - before that step x = 0 was still a valid solution to your equations, but after it it wasn't
Expanding on my reply, here's a different way to continue your own work that would have yielded both solutions, by avoiding any steps that divide by x:
6^ln(6) / 5^ln(5) = x^ln(1.2) → Add right term to right side, divide by its coefficient
(5^(ln(6)/ln(5)))^ln(6) / 5^ln(5) = x^ln(1.2) → Convert numerator of left side to have same base as denominator, using change of base formula: log_5(6) = ln(6)/ln(5)