Thanks very much, and for sure. So this proof applies only to k = \lceil x \log 3 \rceil.
I was intending to use \cfrac{3^x(x/3)}{2^k-3^x}, which @Collag3n posted here, and I tried to recapitulate here.
If I understand right, that minimal-member upper bound more flexibly applies to any k \geq \lceil x \log 3 \rceil, though it less flexibly applies only to integer cycles.