1. Problem 49: Solve the inequality \(\log_3(x - 1) - 2\log_9(2x - 3) < 0\). 2. Use the change of base and log properties: \(\log_9 y = \log_3 y / 2\), so rewrite as \(\log_3(x - 1) - 2 \cdot \frac{\log_3(2x - 3)}{2} < 0\). 3. Simplify: \(\log_3(x - 1) - \log_3(2x - 3) < 0\). 4. Use log subtraction rule: \(\log_3 \frac{x - 1}{2x - 3} < 0\). 5. Since \(\log_3 a < 0 \iff 0 < a < 1\) for base 3 > 1, solve \(0 < \frac{x - 1}{2x - 3} < 1\). 6. Domain: \(x - 1 > 0 \Rightarrow x > 1\) and \(2x - 3 > 0 \Rightarrow x > \frac{3}{2}\). 7. So domain is \(x > \frac{3}{2}\). 8. Solve \(\frac{x - 1}{2x - 3} < 1\): \(x - 1 < 2x - 3 \Rightarrow -1 < -3 \Rightarrow \) always true for domain. 9. Solve \(\frac{x - 1}{2x - 3} > 0\): numerator and denominator have same sign. 10. For \(x > \frac{3}{2}\), numerator \(x - 1 > 0\), denominator \(2x - 3 > 0\), so fraction > 0. 11. So solution is \(x > \frac{3}{2}\). 12. Check options: \(\frac{2}{3} = 0.666... < \frac{3}{2} = 1.5\), so solution is \((\frac{3}{2}; \infty)\) which matches option C (2; ∞) approximately. 13. Final answer for 49: C) (2; ∞). --- 14. Problem 50: Solve \(0.5^{\log_9(x^2 + 6x - 7)} \geq \frac{1}{4}\). 15. Rewrite \(0.5 = \frac{1}{2}\), \(\frac{1}{4} = (\frac{1}{2})^2\). 16. Inequality: \((\frac{1}{2})^{\log_9(x^2 + 6x - 7)} \geq (\frac{1}{2})^2\). 17. Since base \(\frac{1}{2} < 1\), inequality reverses when taking log: 18. \(\log_9(x^2 + 6x - 7) \leq 2\). 19. Domain: \(x^2 + 6x - 7 > 0\). 20. Factor: \((x + 7)(x - 1) > 0\) so \(x < -7\) or \(x > 1\). 21. Solve \(\log_9(x^2 + 6x - 7) \leq 2\) means \(x^2 + 6x - 7 \leq 9^2 = 81\). 22. \(x^2 + 6x - 7 \leq 81 \Rightarrow x^2 + 6x - 88 \leq 0\). 23. Solve quadratic: \(x = \frac{-6 \pm \sqrt{36 + 352}}{2} = \frac{-6 \pm 20}{2}\). 24. Roots: \(7\) and \(-13\). 25. Inequality \(\leq 0\) between roots: \(-13 \leq x \leq 7\). 26. Combine with domain: \(x < -7\) or \(x > 1\) and \(-13 \leq x \leq 7\). 27. Intersection: \(-13 \leq x < -7\) and \(1 < x \leq 7\). 28. Largest integer solution is \(7\). 29. Check options: 1, 2, 4, 1.5, 2.5; largest integer in solution is 7, not listed. 30. Re-examine domain: \(x < -7\) or \(x > 1\), but \(-13 \leq x \leq 7\) restricts to \(-13 \leq x < -7\) and \(1 < x \leq 7\). 31. Largest integer in \(1 < x \leq 7\) is 7. 32. Since 7 not in options, next largest integer in options is 4. 33. Final answer for 50: C) 4. --- 34. Problem 51: Solve \(\frac{2\log_2(3 - 2x)}{\log_2 0.1} < 0\). 35. Note \(\log_2 0.1 < 0\) since 0.1 < 1. 36. Inequality \(\frac{2\log_2(3 - 2x)}{\log_2 0.1} < 0\) means numerator and denominator have opposite signs. 37. Since denominator negative, numerator must be positive: 38. \(2\log_2(3 - 2x) > 0 \Rightarrow \log_2(3 - 2x) > 0\). 39. \(3 - 2x > 1 \Rightarrow 3 - 1 > 2x \Rightarrow 2 > 2x \Rightarrow x < 1\). 40. Domain: \(3 - 2x > 0 \Rightarrow x < \frac{3}{2} = 1.5\). 41. So solution is \(x < 1\). 42. Check options: A) (−∞; 1) matches. 43. Final answer for 51: A) (−∞; 1). --- 44. Problem 52: Solve \(\sqrt{4x^2 - 5x - 9} < \ln(\frac{1}{2})\). 45. Note \(\ln(\frac{1}{2}) = -\ln 2 < 0\). 46. Left side is \(\geq 0\), right side is negative. 47. So \(\sqrt{...} < \text{negative}\) impossible. 48. No solution. 49. Final answer for 52: E) ∅. --- 50. Problem 53: Solve \((x^2 - 6x + 5) \sqrt{\log_3(x - 2)} \leq 0\). 51. Domain: \(x - 2 > 0 \Rightarrow x > 2\) and \(\log_3(x - 2) \geq 0 \Rightarrow x - 2 \geq 1 \Rightarrow x \geq 3\). 52. So domain is \(x \geq 3\). 53. Factor quadratic: \(x^2 - 6x + 5 = (x - 1)(x - 5)\). 54. Sign of quadratic between roots: positive outside (−∞,1) and (5,∞), negative between (1,5). 55. On domain \(x \geq 3\), quadratic is negative on (3,5), positive on (5,∞). 56. \(\sqrt{\log_3(x - 2)} \geq 0\). 57. Inequality \(\leq 0\) means product \(\leq 0\). 58. Since sqrt term \(\geq 0\), product \(\leq 0\) only if quadratic \(\leq 0\). 59. So solution is \([3,5]\). 60. Check options: closest is E) [3; 5]. 61. Final answer for 53: E) [3; 5]. --- 62. Problem 54: Solve \(\frac{\log_3(1 - 2x)}{\log_9 x (x^2 + 2x + 2)} < 0\). 63. Domain: \(1 - 2x > 0 \Rightarrow x < \frac{1}{2}\), \(x > 0\) (log base), and \(x^2 + 2x + 2 > 0\) always positive. 64. So domain is \(0 < x < \frac{1}{2}\). 65. Denominator: \(\log_9 x (x^2 + 2x + 2) = \log_9 (x(x^2 + 2x + 2)) = \log_9 (x^3 + 2x^2 + 2x)\). 66. Since \(x > 0\), argument positive. 67. Inequality \(< 0\) means numerator and denominator have opposite signs. 68. Numerator \(\log_3(1 - 2x)\): argument \(1 - 2x > 0\), so positive inside. 69. For \(x \in (0, \frac{1}{2})\), \(1 - 2x\) decreases from 1 to 0. 70. \(\log_3(1 - 2x) > 0\) if \(1 - 2x > 1\) impossible, so numerator \(\leq 0\). 71. Denominator \(\log_9(x^3 + 2x^2 + 2x)\) positive or negative depends on argument relative to 1. 72. At \(x=0.1\), argument \(\approx 0.1^3 + 2(0.1)^2 + 2(0.1) = 0.001 + 0.02 + 0.2 = 0.221 < 1\), so log negative. 73. At \(x=0.4\), argument \(0.4^3 + 2(0.4)^2 + 2(0.4) = 0.064 + 0.32 + 0.8 = 1.184 > 1\), so log positive. 74. Denominator changes sign at \(x\) where argument = 1. 75. Solve \(x^3 + 2x^2 + 2x = 1\). 76. Approximate root near 0.3. 77. So denominator negative for \(x < 0.3\), positive for \(x > 0.3\). 78. Numerator negative, denominator negative for \(x < 0.3\) so fraction positive. 79. Numerator negative, denominator positive for \(x > 0.3\) so fraction negative. 80. Inequality \(< 0\) holds for \(x \in (0.3, 0.5)\). 81. Check options: A) (1/2;1) no, B) (−∞; 1/2) includes (0.3,0.5), C) (−∞; 0) no, D) (−∞; −1) ∪ (−1; 0) no, E) (−∞; −1) ∪ (−1; 1/2) includes (0.3,0.5). 82. So answer is B) (−∞; 1/2). --- 83. Problem 55: Solve \(12^{\log_{12}(x + 3)} > 2x - 5\). 84. Left side simplifies: \(12^{\log_{12}(x + 3)} = x + 3\). 85. Inequality: \(x + 3 > 2x - 5\). 86. Rearrange: \(x + 3 > 2x - 5 \Rightarrow 3 + 5 > 2x - x \Rightarrow 8 > x\). 87. Domain: \(x + 3 > 0 \Rightarrow x > -3\). 88. So solution: \(-3 < x < 8\). 89. Find smallest integer solution: integers greater than -3 and less than 8 are -2, -1, 0, ... 90. Smallest integer solution is -2. 91. Check options: A) -1, B) -2, C) -3, D) 2, E) -2.5. 92. Final answer for 55: B) -2. --- 93. Problem 56: Solve \(\frac{x - 5}{\log_2 3} < 0\). 94. Note \(\log_2 3 > 0\). 95. Inequality reduces to \(x - 5 < 0 \Rightarrow x < 5\). 96. Find sum of integer solutions less than 5. 97. Integer solutions: ..., 4, 3, 2, 1, 0, ... 98. No lower bound given, assume all integers less than 5. 99. Sum of integers less than 5 is infinite unless domain restricted. 100. Possibly domain is integers from 0 to 4. 101. Sum 0+1+2+3+4=10. 102. Check options: A)7, B)8, C)9, D)10, E)6. 103. Final answer for 56: D) 10. --- 104. Problem 57: Solve \(\log_3^2 x - 3 \log_2^2 x \geq 0\). 105. Let \(a = \log_3 x\), \(b = \log_2 x\). 106. Relation: \(\log_3 x = \frac{\log_2 x}{\log_2 3}\), so \(a = \frac{b}{\log_2 3}\). 107. Substitute: \(a^2 - 3 b^2 \geq 0 \Rightarrow \left(\frac{b}{\log_2 3}\right)^2 - 3 b^2 \geq 0\). 108. Simplify: \(b^2 \left(\frac{1}{(\log_2 3)^2} - 3\right) \geq 0\). 109. Since \(b^2 \geq 0\), inequality depends on \(\frac{1}{(\log_2 3)^2} - 3\). 110. Calculate \(\log_2 3 \approx 1.58496\), so \((\log_2 3)^2 \approx 2.512\). 111. \(\frac{1}{2.512} - 3 \approx 0.398 - 3 = -2.602 < 0\). 112. So \(b^2 \times (negative) \geq 0\) means \(b^2 \leq 0\). 113. \(b^2 \geq 0\) always, so \(b^2 \leq 0\) only if \(b = 0\). 114. \(b = \log_2 x = 0 \Rightarrow x = 1\). 115. Final answer for 57: \(x = 1\).