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$\newcommand\ep{\epsilon}$$\newcommand\de{\delta}$Note that $l(x):=(\ln x)/x$ is decreasing in $x\ge e$. For $\ep\in(0,1/e]$, letting $$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$$ and hence $$x_\ep\ge y=\frac1\ep\,\ln\frac1\ep,$$ where $x_\ep\in[e,\infty)$ is the root of the equation $$l(x_\ep)=\ep.$$ On the other hand, for each real $\de>0$, letting $$z:=z_\ep:=(1+\de)y_\ep$$ we have $$l(y)=\ep\frac{\ln\frac{1+\de}\ep+\ln\ln\frac1\ep}{(1+\de)\ln\frac1\ep}\le\ep$$ for all small enough $\ep>0$ and hence $$x_\ep\le z=\frac{1+\de}\ep\,\ln\frac1\ep.$$