Answer
$$(0,1)$$
Work Step by Step
Let $\ln x>0$, then we have
$$\ln x=-a, \quad a>0.$$
Hence, $$ x=e^{-a}=\dfrac{1}{e^a}$$
Note that the lowest value that $x$ can take on is near $e^{\infty}\approx 0$, excluding zero itself. The largest value is $e^0=1$, excluding $1$ itself (because $a\gt 0$).
So, $\ln x $ is negative for $(0,1)$.
