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.