$ =((\ln())^2 + 2 (\ln())−3>0)$, X is $U(0,1)$

Question

Let $X$ be $U(0,1)$. Then, $ =((\ln())^2 + 2 (\ln())−3>0)$. What is $\beta$? My attempt: Let $Y= \ln(X)$. Then, solving $Y+2Y-3 = (Y+3)(Y-1)>0 \implies Y > 1$ or $Y < -3$. But, $Y = \ln(X)$. Hence, we must have $\ln(X) >1 \implies X > e$, which doesn't work because $x \in (0,1)$. Which leaves us with $P(X<e^{-3}) = \frac{1}{e^{-3}}$. Does this make sense?