1. **State the problem:** Solve the equation $$\ln(x - 3) + \ln(x + 4) = 3 \ln 2$$ for $x$. 2. **Use logarithm properties:** Recall that $$\ln a + \ln b = \ln(ab)$$ and $$k \ln a = \ln(a^k)$$. 3. **Apply these properties:** $$\ln(x - 3) + \ln(x + 4) = \ln((x - 3)(x + 4))$$ $$3 \ln 2 = \ln(2^3) = \ln 8$$ So the equation becomes: $$\ln((x - 3)(x + 4)) = \ln 8$$ 4. **Set the arguments equal:** Since $\ln A = \ln B$ implies $A = B$ (for $A,B > 0$), we have: $$(x - 3)(x + 4) = 8$$ 5. **Expand and simplify:** $$x^2 + 4x - 3x - 12 = 8$$ $$x^2 + x - 12 = 8$$ 6. **Bring all terms to one side:** $$x^2 + x - 12 - 8 = 0$$ $$x^2 + x - 20 = 0$$ 7. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=1$, $c=-20$. Calculate the discriminant: $$\Delta = 1^2 - 4 \times 1 \times (-20) = 1 + 80 = 81$$ So, $$x = \frac{-1 \pm \sqrt{81}}{2} = \frac{-1 \pm 9}{2}$$ 8. **Find the two solutions:** $$x_1 = \frac{-1 + 9}{2} = \frac{8}{2} = 4$$ $$x_2 = \frac{-1 - 9}{2} = \frac{-10}{2} = -5$$ 9. **Check domain restrictions:** The arguments of the logarithms must be positive: $$x - 3 > 0 \Rightarrow x > 3$$ $$x + 4 > 0 \Rightarrow x > -4$$ Only $x=4$ satisfies both conditions. **Final answer:** $$x = 4$$