For $t > 1, \, c > 0$

\begin{alignat}{2}

0 &< t^{-1} &&< t^{c-1} \\

0 &< \int_1^x \frac1t \, \mathrm dt &&< \int_1^x t^{c-1}\, \mathrm dt \quad (x > 1) \\

0 &< \ln x &&< \tfrac1c (x^c - 1) < \tfrac1c x^c \\

0 &< \frac{\ln^a x}{x^b} &&< \tfrac1c x^{ac-b} \quad (a > 0, \, b > 0)

\end{alignat}

Now take $c = \frac{b}{2a} > 0$ so that we have

\[ 0 < \frac{\ln^a x}{x^b} < \tfrac1c x^{-\frac{b}{2}} \]

Finally, take limits as $x \to \infty$.

For your result, set $x = e^y$.