Is there a way to show that the logarithmic function is always below $x^ε$? I tried using derivates of $\log x-x^ε$ and got out that the derivative past where $x = 1$ is less than $0$.
So we know that $(\log x- x^ε)' < 0$ for $x > 1$ which means that the difference between their values is decreasing and thus it makes sense that $x^ε > \log x$ because the value between them will be becoming more negative.
On
With the substitution $x\leftarrow e^{y/\epsilon}$, the desired $\ln x<x^{\epsilon}$ turns into $$\frac y\epsilon <e^y.$$ Recall that the probably most important inequality about the exponential function is $$\tag1 e^t\ge 1+t\qquad \text{for }t\in\Bbb R.$$ We conclude from $(1)$ that $$\begin{align} e^y&=e^{y/2}e^{y/2}\\&\ge(1+\tfrac y2)(1+\tfrac y2)&\text{provided }y\ge -2 \\&=1+y+\frac{y^2}4\\&> 1+y+\frac y\epsilon&\text{provided }y>\frac 4\epsilon\\&>\frac y \epsilon\end{align}$$ Therefore, $$ \ln x<x^\epsilon\qquad \text{for }x>e^{\frac{4}{\epsilon^2}}.$$
We want that $$\epsilon>\frac{\ln{t}}{t},$$ where $t=\ln{x}>0$ and $\epsilon>0$.
It's wrong for $t=e$ and $\epsilon<\frac{1}{e},$ but easy to see that $$\lim_{t\rightarrow+\infty}\frac{\ln{t}}{t}=0$$ because for all $t>1$ we have $$1<t^{\frac{1}{t}}<1+\sqrt{\frac{2}{t}}.$$