1. **State the problem:** Solve the equation $$3x \cdot x \cdot x + 2x \cdot x + 4 \ln(x) = 0$$ for $x$. 2. **Rewrite the equation:** Simplify the powers of $x$: $$3x^3 + 2x^2 + 4 \ln(x) = 0$$ 3. **Important notes:** The natural logarithm $\ln(x)$ is defined only for $x > 0$. So, we consider $x > 0$ only. 4. **Approach:** This is a transcendental equation (polynomial plus logarithm), so exact algebraic solutions are difficult. We can analyze or approximate solutions. 5. **Check behavior:** For $x$ close to 0 (but positive), $\ln(x) \to -\infty$, so $4 \ln(x)$ is very negative, while $3x^3 + 2x^2 \to 0$. So the sum tends to $-\infty$. 6. For large $x$, $3x^3$ dominates and is positive, so the sum is positive. 7. By Intermediate Value Theorem, there is at least one positive root where the function crosses zero. 8. **Try to find approximate root:** Test $x=0.5$: $$3(0.5)^3 + 2(0.5)^2 + 4 \ln(0.5) = 3(0.125) + 2(0.25) + 4(-0.6931) = 0.375 + 0.5 - 2.7724 = -1.8974 < 0$$ 9. Test $x=1$: $$3(1)^3 + 2(1)^2 + 4 \ln(1) = 3 + 2 + 0 = 5 > 0$$ 10. So root lies between 0.5 and 1. 11. Test $x=0.7$: $$3(0.7)^3 + 2(0.7)^2 + 4 \ln(0.7) = 3(0.343) + 2(0.49) + 4(-0.3567) = 1.029 + 0.98 - 1.4268 = 0.582 > 0$$ 12. Test $x=0.6$: $$3(0.6)^3 + 2(0.6)^2 + 4 \ln(0.6) = 3(0.216) + 2(0.36) + 4(-0.5108) = 0.648 + 0.72 - 2.043 = -0.675 < 0$$ 13. Root lies between 0.6 and 0.7. 14. By linear interpolation, approximate root $x \approx 0.6 + \frac{0.675}{0.675 + 0.582} \times 0.1 \approx 0.6 + 0.537 \times 0.1 = 0.6537$. **Final answer:** The solution to the equation is approximately $$x \approx 0.65$$ (to two decimal places).