1. **State the problem:** Solve the equation $$(1 + \ln x)^2 = 0$$ for $x$. 2. **Understand the equation:** The square of a quantity equals zero only if the quantity itself is zero. So, we set: $$1 + \ln x = 0$$ 3. **Isolate the logarithm:** $$\ln x = -1$$ 4. **Rewrite the logarithmic equation in exponential form:** Recall that $\ln x$ means $\log_e x$, so: $$x = e^{-1}$$ 5. **Final answer:** $$x = \frac{1}{e}$$ This is the only solution because the square of the expression is zero only when the expression itself is zero, and $x$ must be positive since the logarithm is defined only for positive $x$.