Dato che: \[
\log_a(x) = x^b
\quad \Rightarrow \quad
\log_a\left(x^b\right) = b\,x^b
\quad \Rightarrow \quad
x^b = a^{b\,x^b}
\quad \Rightarrow \quad
x^b = e^{b\,x^b\log(a)}
\] ossia: \[
x^b e^{-b\,x^b\log(a)} = 1
\quad \Rightarrow \quad
-b\,x^b\log(a)\,e^{-b\,x^b\log(a)} = -b\log(a)
\quad \Rightarrow \quad
y\,e^y = -b\log(a)
\] essendo noto che: \[
y\,e^y \ge -\frac{1}{e} \quad \quad \forall\,y \in \mathbb{R}
\] allora deve essere: \[
\boxed{-b\log(a) \ge -\frac{1}{e}\,}
\] ossia: \[
{\color{blue}{\left(0 < a < 1 \quad \land \quad b \ge \frac{1}{e\log(a)}\right)}}
\quad \vee \quad
{\color{red}{\left(a > 1 \quad \land \quad b \le \frac{1}{e\log(a)}\right)}}
\] \(\quad\quad\quad\quad\)