## Calculus (3rd Edition)

$$(0,1)$$
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 $(0,1)$.