1. **Problem (i): Solve the equation** $$2^{2x-1} - 9 \cdot 2^{x-2} + 1 = 0$$ 2. **Rewrite terms using properties of exponents:** $$2^{2x-1} = 2^{2x} \cdot 2^{-1} = \frac{(2^x)^2}{2}$$ $$2^{x-2} = 2^x \cdot 2^{-2} = \frac{2^x}{4}$$ 3. **Substitute** $y = 2^x$ to simplify: $$\frac{y^2}{2} - 9 \cdot \frac{y}{4} + 1 = 0$$ 4. **Multiply entire equation by 4 to clear denominators:** $$4 \cdot \frac{y^2}{2} - 4 \cdot 9 \cdot \frac{y}{4} + 4 \cdot 1 = 0$$ $$2y^2 - 9y + 4 = 0$$ 5. **Solve quadratic equation:** $$2y^2 - 9y + 4 = 0$$ 6. **Use quadratic formula:** $$y = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2} = \frac{9 \pm \sqrt{81 - 32}}{4} = \frac{9 \pm \sqrt{49}}{4}$$ 7. **Calculate roots:** $$y_1 = \frac{9 + 7}{4} = \frac{16}{4} = 4$$ $$y_2 = \frac{9 - 7}{4} = \frac{2}{4} = \frac{1}{2}$$ 8. **Recall substitution $y = 2^x$ and solve for $x$:** $$2^x = 4 \Rightarrow x = 2$$ $$2^x = \frac{1}{2} = 2^{-1} \Rightarrow x = -1$$ --- 9. **Problem (ii): Solve the equation** $$\log_x 4 = 2 + 4 \log_4 2$$ 10. **Recall that** $\log_4 2 = \frac{1}{2}$ because $4^{1/2} = 2$ 11. **Substitute:** $$\log_x 4 = 2 + 4 \cdot \frac{1}{2} = 2 + 2 = 4$$ 12. **Rewrite logarithm in exponential form:** $$x^4 = 4$$ 13. **Solve for $x$:** $$x = \sqrt[4]{4} = 4^{1/4} = (2^2)^{1/4} = 2^{1/2} = \sqrt{2}$$ --- 14. **Problem (b): Solve** $$5 + 4\sqrt{k} = 9 + 8\sqrt{2}$$ 15. **Isolate $\sqrt{k}$:** $$4\sqrt{k} = 9 + 8\sqrt{2} - 5 = 4 + 8\sqrt{2}$$ 16. **Divide both sides by 4:** $$\sqrt{k} = \frac{4 + 8\sqrt{2}}{4} = 1 + 2\sqrt{2}$$ 17. **Square both sides to find $k$:** $$k = (1 + 2\sqrt{2})^2 = 1^2 + 2 \cdot 1 \cdot 2\sqrt{2} + (2\sqrt{2})^2 = 1 + 4\sqrt{2} + 8 = 9 + 4\sqrt{2}$$ 18. **Answer in the form $a + b\sqrt{2}$:** $$k = 9 + 4\sqrt{2}$$ **Final answers:** (i) $x = 2$ or $x = -1$ (ii) $x = \sqrt{2}$ (b) $k = 9 + 4\sqrt{2}$