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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.

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    @Berci Yes. $\phantom{.}$2012-11-23

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