1. **State the problem:** Solve the inequation $$(2e^x - 1)(e^x - 3) \geq 0.$$\n\n2. **Understand the problem:** We want to find all values of $x$ such that the product of the two factors $2e^x - 1$ and $e^x - 3$ is greater than or equal to zero. This means the product is either positive or zero.\n\n3. **Key rule:** A product of two expressions is $\geq 0$ if both expressions are positive or both are negative, or if either is zero.\n\n4. **Find zeros of each factor:**\n- For $2e^x - 1 = 0$, solve for $x$: $$2e^x = 1 \implies e^x = \frac{1}{2} \implies x = \ln\left(\frac{1}{2}\right) = -\ln(2).$$\n- For $e^x - 3 = 0$, solve for $x$: $$e^x = 3 \implies x = \ln(3).$$\n\n5. **Determine sign intervals:** The critical points split the real line into three intervals: $$(-\infty, -\ln(2)), \quad [-\ln(2), \ln(3)], \quad (\ln(3), \infty).$$\n\n6. **Test each interval:**\n- For $x < -\ln(2)$, pick $x = -1$ (since $-\ln(2) \approx -0.693$):\n $$e^{-1} \approx 0.367.$$\n Then, $2e^x - 1 = 2(0.367) - 1 = 0.734 - 1 = -0.266 < 0,$\n and $e^x - 3 = 0.367 - 3 = -2.633 < 0.$\n Product is negative $\times$ negative = positive $\geq 0$.\n\n- For $-\ln(2) < x < \ln(3)$, pick $x=0$: \n $$e^0 = 1.$$\n Then, $2(1) - 1 = 1 > 0,$\n and $1 - 3 = -2 < 0.$\n Product is positive $\times$ negative = negative $< 0$.\n\n- For $x > \ln(3)$, pick $x=2$: \n $$e^2 \approx 7.389.$$\n Then, $2(7.389) - 1 = 14.778 - 1 = 13.778 > 0,$\n and $7.389 - 3 = 4.389 > 0.$\n Product is positive $\times$ positive = positive $\geq 0$.\n\n7. **Include points where product is zero:**\n- At $x = -\ln(2)$, $2e^x - 1 = 0$, so product is zero.\n- At $x = \ln(3)$, $e^x - 3 = 0$, so product is zero.\n\n8. **Final solution set:**\n$$(-\infty, -\ln(2)] \cup [\ln(3), \infty).$$