1. The problem is to solve the inequality $\ln x \geq -1$. 2. Recall that the natural logarithm function $\ln x$ is defined only for $x > 0$. 3. To solve the inequality, we exponentiate both sides to remove the logarithm, using the fact that the exponential function $e^y$ is the inverse of $\ln y$. 4. Applying $e^{(\cdot)}$ to both sides gives: $$e^{\ln x} \geq e^{-1}$$ 5. Since $e^{\ln x} = x$ for $x > 0$, the inequality becomes: $$x \geq e^{-1}$$ 6. Note that $e^{-1} = \frac{1}{e}$, so the solution is: $$x \geq \frac{1}{e}$$ 7. Also remember the domain restriction $x > 0$, which is satisfied by $x \geq \frac{1}{e}$. Final answer: $$x \geq \frac{1}{e}$$