Big-O complexity upon taking exponent
The statement $X = O(\log n)$ gives an upper bound on $X$, which implies a lower bound on $e^{-X}$ (which is a monotone decreasing function). You can say
$$
e^{-X} = e^{-O(\log n)} = n^{-O(1)}.
$$
In other words, there exists a constant $C &gt; 0$ such that for large $n$,
$$ e^{-X} \geq \frac{1}{n^C}. $$