1. **State the problem:** Solve the inequality $$2\left(\frac{1+\ln(x)}{x}\right) > 0$$ for $x$. 2. **Rewrite the inequality:** Since 2 is positive, the inequality depends on $$\frac{1+\ln(x)}{x} > 0$$. 3. **Analyze the fraction:** A fraction is positive if numerator and denominator have the same sign. So either: - $1+\ln(x) > 0$ and $x > 0$, or - $1+\ln(x) < 0$ and $x < 0$. 4. **Domain restriction:** Since $\ln(x)$ is defined only for $x > 0$, the second case $x < 0$ is invalid. So we only consider $x > 0$. 5. **Solve numerator inequality:** $$1 + \ln(x) > 0 \implies \ln(x) > -1$$ Exponentiate both sides: $$x > e^{-1} = \frac{1}{e}$$ 6. **Combine domain and numerator condition:** Since $x > 0$ and $x > \frac{1}{e}$, the solution is $$x > \frac{1}{e}$$ 7. **Final answer:** $$\boxed{x > \frac{1}{e}}$$