I would like to pick $x$ as small as possible while guaranteeing that $\log(x)/x \leq \epsilon$ where $\epsilon \ll 1$. Clearly $x$ should (roughly) be of the order $1/\epsilon$; I would like simple upper and lower bounds on how big $x$ can be that don't rely on solving transcendental equations.
Edit: It seems to me that taking $x = \epsilon^{-1}\log(\epsilon^{-(1 + \delta)})$ should satisfy the bound for any $\delta > 0$, provided $\epsilon$ is sufficiently small.
$\newcommand\ep{\epsilon}$$\newcommand\de{\delta}$We shall be assuming that $\ep\in(0,1/e]$. Note that $l(x):=(\ln x)/x$ is decreasing in $x\ge e$. So, for $x\ge e$ we have $$(\ln x)/x\le\ep\iff x\ge x_\ep,$$ where $x_\ep\in[e,\infty)$ is the root of the equation $$l(x_\ep)=\ep.$$ Letting $$y:=y_\ep:=\frac1\ep\,\ln\frac1\ep\ge e,$$ we have $$l(y)=\ep\frac{\ln\frac1\ep+\ln\ln\frac1\ep}{\ln\frac1\ep}\ge\ep=l(x_\ep)$$ and hence $$x_\ep\ge y=\frac1\ep\,\ln\frac1\ep.\tag{1}$$
On the other hand, for each real $\de>0$, letting $$z:=z_\ep:=(1+\de)y_\ep$$ we have $$l(z)=\ep\frac{\ln\frac{1+\de}\ep+\ln\ln\frac1\ep}{(1+\de)\ln\frac1\ep}\le\ep=l(x_\ep)$$ for all small enough $\ep>0$ and hence $$x_\ep\le z=\frac{1+\de}\ep\,\ln\frac1\ep.\tag{2}$$
In particular, it follows that $$x_\ep\sim\frac1\ep\,\ln\frac1\ep$$ as $\ep\downarrow0$.
Working similarly but just a bit harder, we can see that the upper bound in (2) on $x_\ep$ will hold if, instead of taking a constant $\de>0$, we take $$\de=c\eta,\quad\text{with}\quad \eta:=\frac{\ln\ln\frac1\ep}{\ln\frac1\ep},$$ for any fixed real $c>1$ and all small enough $\ep>0$. On the other hand, the lower bound on $x_\ep$ in (1) can be improved as follows: $$x_\ep\ge \frac{1+\eta}\ep\,\ln\frac1\ep$$ for all real $x\ge e$.
If you want the smallest, try $$x = -LambertW(-\epsilon)/\epsilon = 1+\epsilon+{\frac{3}{2}}{\epsilon}^{2}+{\frac{8}{3}}{\epsilon}^{3}+{\frac{125}{24}}{\epsilon}^{4} +O \left( {\epsilon}^{5} \right) $$
