1. **Solve** $2 \log_3(4x + 5) - \log_3(x + 3) = 2$. Step 1: State the problem. We need to solve for $x$ in the logarithmic equation. Step 2: Use logarithm properties. Recall that $a \log_b c = \log_b c^a$ and $\log_b m - \log_b n = \log_b \frac{m}{n}$. Step 3: Apply these properties. $$2 \log_3(4x + 5) - \log_3(x + 3) = \log_3( (4x + 5)^2 ) - \log_3(x + 3) = \log_3 \left( \frac{(4x + 5)^2}{x + 3} \right)$$ Step 4: Set the equation equal to 2. $$\log_3 \left( \frac{(4x + 5)^2}{x + 3} \right) = 2$$ Step 5: Convert logarithmic to exponential form. $$\frac{(4x + 5)^2}{x + 3} = 3^2 = 9$$ Step 6: Multiply both sides by $x + 3$. $$(4x + 5)^2 = 9(x + 3)$$ Step 7: Expand and simplify. $$(4x + 5)^2 = 16x^2 + 40x + 25$$ So, $$16x^2 + 40x + 25 = 9x + 27$$ Step 8: Bring all terms to one side. $$16x^2 + 40x + 25 - 9x - 27 = 0$$ $$16x^2 + 31x - 2 = 0$$ Step 9: Solve quadratic equation using the quadratic formula: $$x = \frac{-31 \pm \sqrt{31^2 - 4 \times 16 \times (-2)}}{2 \times 16} = \frac{-31 \pm \sqrt{961 + 128}}{32} = \frac{-31 \pm \sqrt{1089}}{32}$$ Step 10: Calculate the square root. $$\sqrt{1089} = 33$$ Step 11: Find the two solutions. $$x_1 = \frac{-31 + 33}{32} = \frac{2}{32} = \frac{1}{16}$$ $$x_2 = \frac{-31 - 33}{32} = \frac{-64}{32} = -2$$ Step 12: Check domain restrictions. Arguments of logarithms must be positive: $$4x + 5 > 0 \Rightarrow x > -\frac{5}{4}$$ $$x + 3 > 0 \Rightarrow x > -3$$ Both conditions imply $x > -\frac{5}{4}$. Step 13: Check solutions against domain. $x_1 = \frac{1}{16} > -\frac{5}{4}$ valid. $x_2 = -2 < -\frac{5}{4}$ invalid. **Final solution for (i):** $$x = \frac{1}{16}$$ --- 2. Given $a > 0$, $b > 0$ and $$\log_{10} a + \log_{10} b = \log_{10} (a + b)$$ (a) **Prove that** $$a = \frac{b}{b - 1}$$ Step 1: Use logarithm addition property. $$\log_{10} a + \log_{10} b = \log_{10} (ab)$$ Step 2: Equate the expressions. $$\log_{10} (ab) = \log_{10} (a + b)$$ Step 3: Since logarithms are equal and base 10, their arguments are equal. $$ab = a + b$$ Step 4: Rearrange the equation. $$ab - a - b = 0$$ $$a(b - 1) = b$$ Step 5: Solve for $a$. $$a = \frac{b}{b - 1}$$ **Proof complete.** (b) **Write down the full restriction on $b$ with reason.** Step 1: Recall domain restrictions. Since $a > 0$ and $b > 0$, and $a = \frac{b}{b - 1}$, we require: - Denominator $b - 1 \neq 0 \Rightarrow b \neq 1$ - $a > 0 \Rightarrow \frac{b}{b - 1} > 0$ Step 2: Analyze the inequality $\frac{b}{b - 1} > 0$. - If $b > 1$, then $b - 1 > 0$, so fraction is positive. - If $0 < b < 1$, then $b > 0$ but $b - 1 < 0$, so fraction is negative. Step 3: Combine restrictions. $$b > 0, b \neq 1, \text{ and } b > 1$$ Therefore, $$b > 1$$ **Final restriction:** $$b > 1$$ --- **Summary:** (i) $x = \frac{1}{16}$ (ii)(a) $a = \frac{b}{b - 1}$ (ii)(b) $b > 1$