1. Problem Q4: Evaluate logarithms. 2. Recall the logarithm definition: $\log_b a = c$ means $b^c = a$. 3. a) $\log_4 64$: Find $c$ such that $4^c = 64$. 4. Express 64 as a power of 4: $64 = 4^3$ since $4^3 = 64$. 5. So, $\log_4 64 = 3$. 6. b) $\log_{\sqrt{3}} 27$: Find $c$ such that $(\sqrt{3})^c = 27$. 7. Write bases as powers of 3: $\sqrt{3} = 3^{1/2}$ and $27 = 3^3$. 8. So, $(3^{1/2})^c = 3^3$ implies $3^{c/2} = 3^3$. 9. Equate exponents: $\frac{c}{2} = 3 \Rightarrow c = 6$. 10. Thus, $\log_{\sqrt{3}} 27 = 6$. 11. c) $\log_9 3$: Find $c$ such that $9^c = 3$. 12. Express 9 and 3 as powers of 3: $9 = 3^2$, so $(3^2)^c = 3$. 13. This gives $3^{2c} = 3^1$. 14. Equate exponents: $2c = 1 \Rightarrow c = \frac{1}{2}$. 15. So, $\log_9 3 = \frac{1}{2}$. 16. Problem Q5: Solve for $x$ in logarithmic equations. 17. a) $\log_2 x = 5$: means $2^5 = x$. 18. Calculate $2^5 = 32$. 19. So, $x = 32$. 20. b) $\log_x 49 = 2$: means $x^2 = 49$. 21. Solve $x^2 = 49$ gives $x = \pm 7$. 22. Since base of log must be positive and not 1, $x = 7$. 23. c) $\log_3 (x - 2) = 2$: means $3^2 = x - 2$. 24. Calculate $3^2 = 9$. 25. So, $x - 2 = 9 \Rightarrow x = 11$. 26. Problem Q6: Simplify logarithmic expressions. 27. a) $\log a + \log b - \log ab$. 28. Use log rules: $\log a + \log b = \log (ab)$. 29. So expression becomes $\log (ab) - \log (ab)$. 30. Using $\log m - \log n = \log \frac{m}{n}$, we get $\log \frac{ab}{ab} = \log 1$. 31. Since $\log 1 = 0$, the expression simplifies to 0. 32. b) $\log_5 125^{2x} - \frac{1}{2} \log_5 25$. 33. Express 125 and 25 as powers of 5: $125 = 5^3$, $25 = 5^2$. 34. So, $\log_5 (5^3)^{2x} - \frac{1}{2} \log_5 (5^2)$. 35. Simplify exponents: $\log_5 5^{6x} - \frac{1}{2} \log_5 5^2$. 36. Use $\log_b b^k = k$: $6x - \frac{1}{2} \times 2 = 6x - 1$. 37. So, the simplified expression is $6x - 1$. 38. Problem Q7: Solve $\log(3x - 1) + \log(x - 2) = \log 14$. 39. Use log addition rule: $\log A + \log B = \log (AB)$. 40. So, $\log[(3x - 1)(x - 2)] = \log 14$. 41. Equate arguments: $(3x - 1)(x - 2) = 14$. 42. Expand: $3x^2 - 6x - x + 2 = 14$. 43. Simplify: $3x^2 - 7x + 2 = 14$. 44. Subtract 14: $3x^2 - 7x + 2 - 14 = 0 \Rightarrow 3x^2 - 7x - 12 = 0$. 45. Solve quadratic: $3x^2 - 7x - 12 = 0$. 46. Use quadratic formula: $x = \frac{7 \pm \sqrt{(-7)^2 - 4 \times 3 \times (-12)}}{2 \times 3}$. 47. Calculate discriminant: $49 + 144 = 193$. 48. So, $x = \frac{7 \pm \sqrt{193}}{6}$. 49. Check domain: $3x - 1 > 0 \Rightarrow x > \frac{1}{3}$ and $x - 2 > 0 \Rightarrow x > 2$. 50. So, $x > 2$. 51. Approximate roots: $\sqrt{193} \approx 13.89$. 52. $x_1 = \frac{7 + 13.89}{6} = \frac{20.89}{6} \approx 3.48$ (valid). 53. $x_2 = \frac{7 - 13.89}{6} = \frac{-6.89}{6} \approx -1.15$ (invalid). 54. Final solution: $x \approx 3.48$. 55. Problem Q8: Solve $2^{3x} + 2^{2x} = 96$. 56. Let $y = 2^x$. 57. Then $2^{3x} = (2^x)^3 = y^3$ and $2^{2x} = y^2$. 58. Equation becomes $y^3 + y^2 = 96$. 59. Rearrange: $y^3 + y^2 - 96 = 0$. 60. Try integer roots: test $y=4$. 61. Calculate $4^3 + 4^2 = 64 + 16 = 80$ (not 96). 62. Test $y=3$: $27 + 9 = 36$ (no). 63. Test $y=2$: $8 + 4 = 12$ (no). 64. Test $y=6$: $216 + 36 = 252$ (no). 65. Test $y=3.5$: $42.875 + 12.25 = 55.125$ (no). 66. Test $y=4.5$: $91.125 + 20.25 = 111.375$ (no). 67. Use substitution $y^2(y + 1) = 96$. 68. Try $y=4$: $16 \times 5 = 80$ (close). 69. Try $y=4.2$: $17.64 \times 5.2 = 91.73$ (closer). 70. Try $y=4.3$: $18.49 \times 5.3 = 98.0$ (too high). 71. So root is between 4.2 and 4.3. 72. Use quadratic in $y$: rewrite as $y^3 + y^2 - 96 = 0$. 73. Use numerical methods or approximate root $y \approx 4.25$. 74. Since $y = 2^x$, $2^x = 4.25$. 75. Take log base 2: $x = \log_2 4.25$. 76. Calculate $x = \frac{\log 4.25}{\log 2} \approx \frac{0.628}{0.301} \approx 2.09$. Final answers: Q4: a) 3, b) 6, c) 1/2 Q5: a) 32, b) 7, c) 11 Q6: a) 0, b) $6x - 1$ Q7: $x \approx 3.48$ Q8: $x \approx 2.09$