Yes; see, e.g., the answer at https://mathoverflow.net/a/116852 https://mathoverflow.net/a/116852, adjacent to the one cited by the parent. This gives a hard lo

Yes; see, e.g., the answer at https://mathoverflow.net/a/116852 https://mathoverflow.net/a/116852, adjacent to the one cited by the parent. This gives a hard lower bound of |2^x - 3^y| ≥ max(2^x, 3^y)/((e max(x, y))^821013301). (But it's not sufficient for the parent's argument, since (3 + 1/n₁)^o = 3^o·(1 + 1/(3n₁))^o grows larger than 3^o by an exponential factor, so we can always increase (e, o) until the interval is wide enough.)