1. **State the problem:** Given the equation $$3 \log_3(2x - 1) = 2 + \log_3(14x - 25)$$, we need to show that it leads to the cubic equation $$2x^3 - 3x^2 - 30x + 56 = 0$$. 2. **Use logarithm properties:** Recall that $$a \log_b(c) = \log_b(c^a)$$ and $$\log_b(m) + \log_b(n) = \log_b(mn)$$. 3. **Rewrite the left side:** $$3 \log_3(2x - 1) = \log_3((2x - 1)^3)$$. 4. **Rewrite the right side:** $$2 + \log_3(14x - 25) = \log_3(3^2) + \log_3(14x - 25) = \log_3(9) + \log_3(14x - 25) = \log_3(9(14x - 25))$$. 5. **Set the logarithms equal:** Since $$\log_3((2x - 1)^3) = \log_3(9(14x - 25))$$, their arguments must be equal: $$ (2x - 1)^3 = 9(14x - 25) $$ 6. **Expand the left side:** $$(2x - 1)^3 = (2x - 1)(2x - 1)(2x - 1)$$ First, $$(2x - 1)^2 = 4x^2 - 4x + 1$$ Then multiply by $$(2x - 1)$$: $$ (4x^2 - 4x + 1)(2x - 1) = 8x^3 - 8x^2 + 2x - 4x^2 + 4x - 1 = 8x^3 - 12x^2 + 6x - 1 $$ 7. **Rewrite the equation:** $$8x^3 - 12x^2 + 6x - 1 = 9(14x - 25) = 126x - 225$$ 8. **Bring all terms to one side:** $$8x^3 - 12x^2 + 6x - 1 - 126x + 225 = 0$$ Simplify: $$8x^3 - 12x^2 - 120x + 224 = 0$$ 9. **Divide entire equation by 4:** $$2x^3 - 3x^2 - 30x + 56 = 0$$ This matches the required cubic equation. --- **(b) Show that -4 is a root:** 1. Substitute $$x = -4$$ into the cubic: $$2(-4)^3 - 3(-4)^2 - 30(-4) + 56 = 2(-64) - 3(16) + 120 + 56 = -128 - 48 + 120 + 56$$ 2. Calculate stepwise: $$-128 - 48 = -176$$ $$-176 + 120 = -56$$ $$-56 + 56 = 0$$ Since the result is 0, $$x = -4$$ is a root. --- **(c) Solve the original equation:** 1. Since $$x = -4$$ is a root, factor the cubic by dividing by $$x + 4$$. 2. Use polynomial division or synthetic division: Divide $$2x^3 - 3x^2 - 30x + 56$$ by $$x + 4$$. 3. Synthetic division: - Coefficients: 2, -3, -30, 56 - Root to test: -4 Carry down 2. Multiply: 2 * (-4) = -8. Add: -3 + (-8) = -11. Multiply: -11 * (-4) = 44. Add: -30 + 44 = 14. Multiply: 14 * (-4) = -56. Add: 56 + (-56) = 0. Quotient polynomial: $$2x^2 - 11x + 14$$ 4. Solve quadratic $$2x^2 - 11x + 14 = 0$$ using the quadratic formula: $$x = \frac{11 \pm \sqrt{(-11)^2 - 4 \times 2 \times 14}}{2 \times 2} = \frac{11 \pm \sqrt{121 - 112}}{4} = \frac{11 \pm \sqrt{9}}{4}$$ 5. Calculate roots: $$x = \frac{11 + 3}{4} = \frac{14}{4} = 3.5$$ $$x = \frac{11 - 3}{4} = \frac{8}{4} = 2$$ 6. Check domain restrictions for original logs: - $$2x - 1 > 0 \Rightarrow x > 0.5$$ - $$14x - 25 > 0 \Rightarrow x > \frac{25}{14} \approx 1.7857$$ So valid solutions are $$x = 2$$ and $$x = 3.5$$. **Final solutions:** $$x = 2$$ and $$x = 3.5$$. --- **Summary:** - Cubic equation derived: $$2x^3 - 3x^2 - 30x + 56 = 0$$ - Verified root: $$x = -4$$ - Other roots found: $$x = 2, 3.5$$ - Valid solutions to original equation: $$x = 2, 3.5$$