If you want to have rationals and keep k \geq \lceil x \log 3 \rceil you can still use this:
From this post we have a_{min}\leqslant \cfrac{1}{2^{k/x}-3}, and with 2^{k/x}>3 \Rightarrow \cfrac{2^k-3^x}{2^\frac{k}{x}-3}=(2^\frac{k}{x})^{x-1}\cdot 3^0+(2^\frac{k}{x})^{x-2}\cdot 3^1+...+(2^\frac{k}{x})^0\cdot 3^{x-1}<x(2^\frac{k}{x})^{x-1}
you have a_{min}\leqslant \cfrac{x(2^\frac{k}{x})^{x-1}}{2^k-3^x} instead of the “integer” bound a_{min}\leqslant \cfrac{x(3)^{x-1}}{2^k-3^x}