1. **Problem 9 (i): Solve the equation** $$2^{2x} - 9(2^x) + 8 = 0$$ 2. **Rewrite using substitution:** Let $$y = 2^x$$. Then $$2^{2x} = (2^x)^2 = y^2$$. 3. **Rewrite the equation:** $$y^2 - 9y + 8 = 0$$ 4. **Factor the quadratic:** $$y^2 - 9y + 8 = (y - 8)(y - 1) = 0$$ 5. **Solve for y:** - $$y - 8 = 0 \Rightarrow y = 8$$ - $$y - 1 = 0 \Rightarrow y = 1$$ 6. **Recall substitution:** $$y = 2^x$$, so - $$2^x = 8 = 2^3 \Rightarrow x = 3$$ - $$2^x = 1 = 2^0 \Rightarrow x = 0$$ --- 7. **Problem 9 (ii): Solve the equation** $$3^{2x} - 10(3^x) + 9 = 0$$ 8. **Substitute:** Let $$z = 3^x$$, then $$3^{2x} = z^2$$. 9. **Rewrite the equation:** $$z^2 - 10z + 9 = 0$$ 10. **Factor the quadratic:** $$z^2 - 10z + 9 = (z - 9)(z - 1) = 0$$ 11. **Solve for z:** - $$z - 9 = 0 \Rightarrow z = 9$$ - $$z - 1 = 0 \Rightarrow z = 1$$ 12. **Recall substitution:** $$z = 3^x$$, so - $$3^x = 9 = 3^2 \Rightarrow x = 2$$ - $$3^x = 1 = 3^0 \Rightarrow x = 0$$ --- 13. **Problem 10: Express powers of 2 in terms of $$y = 2^x$$:** (i) $$2^{2x} = (2^x)^2 = y^2$$ (ii) $$2^{2x+1} = 2^{2x} \cdot 2^1 = y^2 \cdot 2 = 2y^2$$ (iii) $$2^{x+3} = 2^x \cdot 2^3 = y \cdot 8 = 8y$$ --- 14. **Solve the equation:** $$2^{2x+1} - 2^{x+3} - 2^x + 4 = 0$$ 15. **Rewrite using substitution:** $$2y^2 - 8y - y + 4 = 0$$ 16. **Simplify:** $$2y^2 - 9y + 4 = 0$$ 17. **Factor the quadratic:** $$2y^2 - 9y + 4 = (2y - 1)(y - 4) = 0$$ 18. **Solve for y:** - $$2y - 1 = 0 \Rightarrow y = \frac{1}{2}$$ - $$y - 4 = 0 \Rightarrow y = 4$$ 19. **Recall substitution:** $$y = 2^x$$, so - $$2^x = \frac{1}{2} = 2^{-1} \Rightarrow x = -1$$ - $$2^x = 4 = 2^2 \Rightarrow x = 2$$ **Final answers:** - Problem 9 (i): $$x = 0, 3$$ - Problem 9 (ii): $$x = 0, 2$$ - Problem 10: $$x = -1, 2$$