Is it true that $X(t)^a > K \iff X(t) > K^\frac1a$

Question

Let $a \in \mathbb{N}$, $K \in \mathbb{R^+}$ and $X(t)$ be a geometric Brownian Motion. Is the following true?

$$X(t)^a > K \iff X(t) > K^\frac1a$$

The context of the above is that I want to evaluate the probability that $X(t)^a > K$, and this transformation of the inequality would be helpful.

Answer 1

Yes. The function $\mathbb R^+_0 \to \mathbb R^+_0$, $x \mapsto x^{1/a}$ is strictly monotone.