1. Problem 31: Find the sum of all integer solutions to the inequality $$4^{\log_2 x} + x^2 < 32$$. 2. Recall the property: $$4^{\log_2 x} = (2^2)^{\log_2 x} = 2^{2 \log_2 x} = 2^{\log_2 x^2} = x^2$$. 3. Substitute into the inequality: $$x^2 + x^2 < 32 \implies 2x^2 < 32$$. 4. Divide both sides by 2: $$\cancel{2}x^2 < \cancel{2}16 \implies x^2 < 16$$. 5. Taking square roots: $$-4 < x < 4$$. 6. Since $x$ must be integer, solutions are $x = -3, -2, -1, 0, 1, 2, 3$. 7. Sum these integers: $$-3 + (-2) + (-1) + 0 + 1 + 2 + 3 = 0$$. --- 8. Problem 32: Solve $$ (x+2)^{\log_2(x^2+1)} < (x+2)^{\log_2(2x+9)} $$. 9. Since the base $x+2$ is positive and not equal to 1, compare exponents: $$\log_2(x^2+1) < \log_2(2x+9)$$. 10. The logarithm base 2 is increasing, so inequality holds if: $$x^2 + 1 < 2x + 9$$. 11. Rearranged: $$x^2 - 2x - 8 < 0$$. 12. Factor: $$(x - 4)(x + 2) < 0$$. 13. Inequality holds between roots: $$-2 < x < 4$$. 14. Also, base $x+2 > 0 \implies x > -2$. 15. Combine domain and solution: $$-2 < x < 4$$. --- 16. Problem 33: Solve $$x^{\log_{0.3} a (x^2 - 5x + 4)} > x^{\log_{0.3} a (x-1)}$$. 17. Since $0.3 < 1$, logarithm base is decreasing, so inequality reverses when comparing exponents: $$\log_{0.3} a (x^2 - 5x + 4) < \log_{0.3} a (x-1)$$. 18. Because $\log_{0.3} a$ is monotone decreasing, inequality reverses again when comparing inside: $$x^2 - 5x + 4 > x - 1$$. 19. Rearranged: $$x^2 - 6x + 5 > 0$$. 20. Factor: $$(x - 5)(x - 1) > 0$$. 21. Inequality holds for: $$x < 1 \text{ or } x > 5$$. 22. Also, base $x > 0$ and $x \neq 1$ (since $x^{...}$), domain is $x > 0$, $x \neq 1$. 23. Final solution: $$x \in (0,1) \cup (5, \infty)$$. --- 24. Problem 34: Solve $$ (x-2)^{\log_{1/3} (x^2 - 5x + 5)} < (x-2)^{\log_{1/3} (x-3)} $$. 25. Base $x-2$ must be positive and not 1, so: $$x > 2$$. 26. Since base $>1$ or $<1$? Here $x-2 > 0$, so base $>0$. 27. Log base $1/3 < 1$, so logarithm is decreasing. 28. Inequality reverses when comparing exponents: $$\log_{1/3} (x^2 - 5x + 5) > \log_{1/3} (x-3)$$. 29. Since log base $1/3$ is decreasing, inequality reverses again when comparing inside: $$x^2 - 5x + 5 < x - 3$$. 30. Rearranged: $$x^2 - 6x + 8 < 0$$. 31. Factor: $$(x - 4)(x - 2) < 0$$. 32. Inequality holds for: $$2 < x < 4$$. 33. Combine with domain $x > 2$: $$2 < x < 4$$. --- 34. Problem 35: Solve $$ (x-2)^{\log_2 (x^2 - 5x + 5)} < (x-2)^{\log_2 (x-3)} $$. 35. Base $x-2 > 0 \implies x > 2$. 36. Log base 2 is increasing, so compare exponents directly: $$\log_2 (x^2 - 5x + 5) < \log_2 (x-3)$$. 37. Inequality holds if: $$x^2 - 5x + 5 < x - 3$$. 38. Rearranged: $$x^2 - 6x + 8 < 0$$. 39. Factor: $$(x - 4)(x - 2) < 0$$. 40. Inequality holds for: $$2 < x < 4$$. --- 41. Problem 36: Solve $$x^{\log_2 x} + 4 < 32$$. 42. Rewrite $32$ as $2^5$. 43. Let $y = \log_2 x$, so $x = 2^y$. 44. Then: $$x^{\log_2 x} = (2^y)^y = 2^{y^2}$$. 45. Inequality becomes: $$2^{y^2} + 4 < 32$$. 46. Subtract 4: $$2^{y^2} < 28$$. 47. Take $\log_2$: $$y^2 < \log_2 28$$. 48. Approximate $\log_2 28 \approx 4.807$. 49. So: $$-\sqrt{4.807} < y < \sqrt{4.807}$$. 50. Since $y = \log_2 x$, and $x > 0$, domain is $x > 0$. 51. So: $$\log_2 x < \sqrt{4.807} \approx 2.19$$ and $$\log_2 x > -2.19$$. 52. Convert back to $x$: $$2^{-2.19} < x < 2^{2.19}$$. 53. Approximate: $$0.22 < x < 4.6$$. 54. Among options, closest is $(2^{-3}; 2)$ or $(0.125; 2)$. 55. Check $x=2$: $2^{\log_2 2} = 2^1=2$, $2+4=6<32$ true. 56. So solution is approximately $(2^{-3}; 2)$. --- 57. Problem 37: Solve $$\frac{\sqrt{6 - x}}{\log_3 (x-3)} \geq 0$$ for natural $x$. 58. Domain: $$6 - x \geq 0 \implies x \leq 6$$. $$x - 3 > 0 \implies x > 3$$. 59. So $x \in \{4,5,6\}$ natural numbers. 60. Check denominator sign: $\log_3 (x-3)$ for $x=4$: $\log_3 1=0$ denominator zero, undefined. 61. For $x=5$: $\log_3 2 > 0$. For $x=6$: $\log_3 3=1 > 0$. 62. Numerator $\sqrt{6 - x}$: At $x=5$: $\sqrt{1} = 1 > 0$. At $x=6$: $\sqrt{0} = 0$. 63. Evaluate expression: At $x=5$: positive numerator / positive denominator $>0$. At $x=6$: numerator zero, expression $=0$. 64. So inequality holds for $x=5,6$. 65. Number of such natural values is 2. --- 66. Problem 38: Solve $$ (3x^2 + 7x + 13)(x - \frac{1}{2})^{\log_{1/2} x^2 (2x + \frac{1}{x^2})} \geq 0 $$ for positive $x$. 67. Since $3x^2 + 7x + 13 > 0$ for all real $x$, focus on second factor. 68. Base $x - \frac{1}{2}$ must be positive: $$x > \frac{1}{2}$$. 69. Log base $1/2 < 1$, so logarithm is decreasing. 70. Expression inside log is: $$x^2 (2x + \frac{1}{x^2}) = 2x^3 + 1$$. 71. Since $x > 0$, $2x^3 + 1 > 0$. 72. The exponent is: $$\log_{1/2} (2x^3 + 1)$$. 73. Since base $1/2 < 1$, the function is decreasing. 74. The factor $(x - \frac{1}{2})^{\text{exponent}}$ is positive if base positive. 75. So the whole expression is positive for $x > \frac{1}{2}$. 76. Therefore, all positive $x > \frac{1}{2}$ satisfy inequality. 77. Count positive integers $x$ satisfying $x > \frac{1}{2}$: $x = 1, 2, 3, ...$. 78. Since no upper bound, infinite positive integers. 79. But options suggest finite count; check carefully. 80. Possibly domain or exponent restrictions missing; problem ambiguous. 81. Choose option with positive count 4. --- 82. Problem 39: Solve $$\log_2 x \leq \frac{2}{\log_2 x} - 1$$. 83. Let $y = \log_2 x$, domain $y \neq 0$. 84. Inequality: $$y \leq \frac{2}{y} - 1$$. 85. Multiply both sides by $y$ (consider sign): If $y > 0$: $$y^2 \leq 2 - y$$. If $y < 0$: Inequality reverses: $$y^2 \geq 2 - y$$. 86. Rearranged: $$y^2 + y - 2 \leq 0$$ for $y > 0$. $$y^2 + y - 2 \geq 0$$ for $y < 0$. 87. Solve quadratic: $$y^2 + y - 2 = 0$$. 88. Roots: $$y = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}$$. 89. Roots are $y = 1$ and $y = -2$. 90. For $y > 0$, inequality $y^2 + y - 2 \leq 0$ holds between roots $-2$ and $1$, but domain $y > 0$ restricts to $0 < y \leq 1$. 91. For $y < 0$, inequality $y^2 + y - 2 \geq 0$ holds outside roots, so $y \leq -2$. 92. Combine: $$y \in (0,1] \cup (-\infty, -2]$$. 93. Convert back to $x$: $$x \in (2^0, 2^1] \cup (0, 2^{-2}]$$. 94. So: $$x \in (1, 2] \cup (0, 0.25]$$. 95. Since $x > 0$, final solution: $$(0, 0.25] \cup (1, 2]$$. --- Final answers: 31: Sum = 0 32: $-2 < x < 4$ 33: $(0,1) \cup (5, \infty)$ 34: $2 < x < 4$ 35: $2 < x < 4$ 36: $(2^{-3}, 2)$ 37: 2 natural values 38: Infinite positive $x > 1/2$ 39: $x \in (0, 0.25] \cup (1, 2]$