13. Logarithm Solutions Part 2

51. Let \(x = 72^{15}~\therefore \log_{10}x = 15\log_{10}72\)

    \(= 15\log_{10}(2^3*3^2) = 15[3\log_{10}2 + 2\log_{10}3]\)

    \(= 15[3*0.301 + 2*0.477] = 15[0.903 + 0.954] = 15 * 1.857\)

    \(= 27.855\)

    So characteristic is \(27\) the number of digits will be \(28\)

52. Here we make use of the fact that number of integers with base \(b\) and characteristics \(n\) is \(b^{n + 1} - b^n.\)

    Given \(b = 5, n = 2,\) therefore number of integers is \(5^3 - 5^2 = 100\)

53. Let \(x = 3^{15}*2^{10}~\therefore \log_{10}x = 15\log_{10}3 + 10\log_{10}2\)

    \(= 15*0.477 + 10*0.301 = 7.155 + 3.01 = 10.165\)

    Since characteritics is \(10\) number of digits will be \(11\)

54. Let \(x = 6^20~\therefore \log_{10}x = 20\log_{10}2*3\) \(= 20[\log_{10}2 + \log_{10}3] = 20[0.301 + 0.477] = 20*0.778 = 15.56\)

    Since characteristics is \(15\) number of digits will be \(16\)

55. Let \(x = 5^{25}~\therefore \log_{10}x = 25\log_{10}\frac{10}{2}\)

    \(= 25[1 - \log_{10}2] = 25*0.699 = 17.475\)

    Since characteristics is \(17\) number of digits would be \(18\)

56. \(\log_a [1 + \log_b \{1 + \log_c (1 + \log_p x)\}] = 0\)

    \(1 + \log_b \{1 + \log_c (1 + \log_p x)\}] = 1\)

    \(\log_b \{1 + \log_c (1 + \log_p x)\}] = 0\)

    \(1 + \log_c (1 + \log_p x) = 1\)

    \(\log_c (1 + \log_p x) = 0\)

    \(1 + \log_p x = 1\)

    \(\log_p x = 0\)

    \(x = 1\)

57. Given, \(\log_7\log_5(\sqrt{x + 5} + \sqrt{x}) = 0\)

    \(\log_5(\sqrt{x + 5} + \sqrt{x}) = 1\)

    \(\sqrt{x + 5} + \sqrt{x} = 5\)

    \(\sqrt{x + 5} = 5 - \sqrt{x}\)

    Squaring both sides, we get

    \(x + 5 = 25 + x -10\sqrt{x}\)

    \(\sqrt{x} = 2~\Rightarrow x = 4\)

58. \(\log_2 x + \log_4 (x + 2) = 2\)

    \(\log_2 x + \log_{2^2} (x + 2) = 2\)

    \(\log_2 x + \frac{1}{2}\log_2 (x + 2) = 2\)

    \(2\log_2 x + \log_2 (x + 2) = 4\)

    \(\log_2 x^2(x + 2) = 4\)

    \(x^2(x + 2) = 16~\Rightarrow x = 2\)

59. \(\log_(x + 2)x + \log_x (x + 2) = \frac{5}{2}\)

    \(\frac{1}{\log_x(x + 2)} + \log_x(x + 1) = \frac{5}{2}\)

    Let \(z = \log_x(x + 2)\)

    \(\therefore \frac{1}{z} + z = \frac{5}{2}\)

    \(2z^2 + 2 - 5z = 0\Rightarrow z = 2, \frac{1}{2}\)

    \(\log_x(x + 2) = 2, \log_x(x + 2) = \frac{1}{2}\)

    \(x + 2 = 2^x, x + 2 = \sqrt{x}\)

    \(x = 2, x^2 - 3x + 4 = 0\Rightarrow x = \frac{3 \pm \sqrt{-7}}{2}\)

    \(\therefore x = 2\)

60. \(\frac{\log (x + 1)}{\log x} = 2\)

    \(\log_x(x + 1) = 2\)

    \(x + 1 = x^2 \Rightarrow x = \frac{1 \pm \sqrt{5}}{2}\)

    \(\because x > 0, x = \frac{1 + \sqrt{5}}{2}\)

61. \(2\log_x a + \log_{ax} a + 3\log_{a^2x} a = 0\)

    \(\frac{2}{\log_a x} + \frac{1}{\log_a ax} + \frac{3}{\log_a a^2x} = 0\)

    \(\frac{2}{\log_a x} + \frac{1}{\log_a a + \log_a x} + \frac{3}{\log_a a^2 + \log_a x} = 0\)

    \(\frac{2}{\log_a x} + \frac{1}{1 + \log_a x} + \frac{3}{2 + \log_a x} = 0\)

    Let \(\log_a x = z,\) so the above equation becomes

    \(\frac{2}{z} + \frac{1}{z + 1} + \frac{3}{z + 2} = 0\)

    \(2(z + 1)(z + 2) + z(z + 2) + 3z(z + 1) = 0\)

    \(2z^2 + 6z + 4 + z^2 + 2z + 3z^2 + 3z = 0\)

    \(6z^2 + 11z + 4 = 0\)

    \(z = \frac{-11 \pm \sqrt{121 - 4.6.4}}{12} = \frac{-11 \pm 5}{12} = \frac{-1}{2}, \frac{-4}{3}\)

    \(\therefore x = a^{\frac{-1}{2}}, a^{\frac{-4}{3}}\)

62. \(x + \log_{10}(1 + 2^x) = x\log_{10}5 + \log_{10}6\)

    \(\log_{10}10^x + \log_{10}(1 + 2^x) = \log_{10}5^x + \log_{10}6\)

    \(\log_{10}10^x(1 + 2^x) = \log_{10}5^x*6\)

    \(10^x(1 + 2^x) = 5^x*6\)

    \(2^x(1 + 2^x) = 2*3\)

    \(2^x = 2, 1 + 2^x = 3~\Rightarrow x = 1\)

63. \(x^{\frac{3}{4}(\log_2 x)^2 + \log_2x - \frac{5}{4}} = \sqrt{2}\)

    Taking \(\log_2\) of both sides

    \(\left(\frac{3}{4}(\log_2 x)^2 + \log_2x - \frac{5}{4}\right)\log_2 x = \frac{1}{2}\log_2 2\)

    \(\left(\frac{3}{4}(\log_2 x)^2 + \log_2x - \frac{5}{4}\right)\log_2 x = \frac{1}{2}\)

    Let \(z = \log_2 x\)

    \(\left(\frac{3}{4}z^2 + z - \frac{5}{4}\right)z = \frac{1}{2}\)

    Solving this cubic equation will lead to \(x = 2, \frac{1}{4}, \frac{1}{\sqrt[3]{2}}\)

64. Given, \((x^2 + 6)^{\log_3 x} = (5x)^{\log_3 x}\)

    \(\log_3x\) has a possible value of \(0\) in that case \(x = 1\)

    If \(\log_3x \neq 0\)

    \(x^2 + 6 = 5x~\Rightarrow x = 2, 3\)

65. Given \((3 + 2\sqrt{2})^{x^2 - 6x + 9} + (3 - 2\sqrt{2})^{x^2 - 6x + 9} = 6\)

    \(3 + 2\sqrt{2} = \frac{1}{3 - 2\sqrt{2}}\)

    \((3 + 2\sqrt{2})^{x^2 - 6x + 9} + (3 + 2\sqrt{2})^{-(x^2 - 6x + 9)} = 6\)

    Let \(z = (3 + 2\sqrt{2})^{x^2 - 6x + 9}\)

    \(z + \frac{1}{z} = 6\)

    \(z = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{3 \pm 2\sqrt{2}}{}\)

    Thus, \(x^2 - 6x + 9 = \pm 1\)

    Thus, \(x = 2, 4\) because other roots are irrational.

66. Given \(\log_8\left(\frac{8}{x^2}\right) \div (\log_8 x)^2 = 3\)

    \(\log_8 8 - \log_8 x^2 = 3 (\log_8 x)^2\)

    \(1 - 2\log_8x = 3(\log_8 x)^2\)

    Let \(z = \log_8 x\)

    \(1 - 2z = 3z^2\)

    \(z = -1, \frac{1}{3}\)

    \(x = 2, \frac{1}{8}\)

67. Given \(\sqrt{\log_2 (x)^4} + 4\log_4\sqrt{\frac{2}{x}} = 2\)

    \(\sqrt{\log_2 (x)^4} + 2\log_2\sqrt{\frac{2}{x}} = 2\)

    \(\sqrt{4\log_2 x} + \log_2 \frac{2}{x} = 2\)

    \(\sqrt{4\log_2 x} + 1 - \log_2 x = 2\)

    Now we can assume \(\log_2 x = z\) and solve leading to \(x = 2\)

68. Given, \(2\log_{10}x - \log_x0.01 = 5\)

    \(2\log_{10}x - \log_x(10)^{-2} = 5\)

    \(2\log_{10}x + 2\log_x{10} = 5\)

    \(2\log_{10}x + \frac{2}{\log_{10}x} = 5\)

    Let \(z = \log_{10}x\)

    \(2z + \frac{2}{z} = 5\)

    \(2z^2 - 5z + 2 = 0\)

    \(z = 2, \frac{1}{2}\)

    \(x = 100, \sqrt{10}\)

69. Given \(\log_{\sin x}2\log_{\cos x}2 + \log_{\sin x} 2 + \log_{\cos x}2 = 0\)

    \(\log_{\sin x}2(\log_{\cos x}2 + 1) + \log_{\cos x}2 = 0\)

    Taking log of both sides with base \(e\):

    \(\frac{\ln 2}{\ln \sin x}\left(\frac {\ln 2}{ln \cos x} +1\right)+\frac {\ln 2}{ln \cos x}=0\)

    \(\frac{1}{\ln \sin x}\left(\frac {\ln 2}{\ln \cos x} +1\right)+\frac {1}{\ln \cos x} = 0\)

    \(\frac{1}{\ln \sin x}\left(\frac {\ln 2}{ln \cos x} +1\right)=-\frac {1}{\ln \cos x}\)

    \(\frac{1}{ln \sin x}(ln 2 +ln \cos x)=-1\)

    \(\ln(\sin 2x) = 1\)

    \(x = 2k\pi + \frac{\pi}{4}, k\in I\)

70. Given \(2^{x + 3} + 2^{x + 2} + 2^{x + 1} = 7^x + 7^{x - 1}\)

    \(2^{x + 1}*7 = 7^{x - 1}* 8\)

    \((x + 1)\log 2 + \log 7 = (x - 1)\log 7 + 3\log 2\)

    Solving this we get \(x = 2\)

71. Given \(\log_{\sqrt{2}\sin x}(1 + \cos x) = 2\)

    \(1 + \cos x = (\sqrt{2}\sin x)^2\)

    \(1 + \cos x = 2\sin^2 x = 2 - 2\cos^2 x\)

    \(2\cos^2 x + \cos x -1 = 0\)

    \(\cos x = \frac{-1\pm \sqrt{1 + 8}}{4} = -1, \frac{1}{2}\)

    \(x = 2n\pi, 2n\pi + \frac{\pi}{3},~n\in I\)

    However, for logarithm to be defined \(\sin x > 0\) therefore \(x = 2n\pi\) is not an acceptable solution.

72. Given \(\log_{10}[198 + \sqrt{x^3 - x^2 - 12x + 36}] = 2\)

    \(98 + \sqrt{x^3 - x^2 - 12x + 36} = 100\)

    \(x^3 - x^2 - 12x + 32 = 0\)

    Solving this we find one appropriate root which is \(x = -4\)

73. Given \(2^x3^{2x} - 100 = 0\)

    \(x\log_{10}2 + 2x\log_{10}3 = 2\)

    Substituting valuues for \(\log_{10}2\) and \(\log_{10}3,\) we get

    \(0.30103x + 0.95424x = 2\)

    \(x = 10593\)

74. Give, \(\log_x 3\log_{\frac{x}{3}}3 + \log{\frac{x}{81}}3 = 0\)

    \(\frac{1}{\log_3 x} + \frac{1}{\log_3\left(\frac{x}{3}\right)} + \frac{1}{\log_3}\left(\frac{x}{81}\right) = 0\)

    \(\frac{1}{\log_3 x}.\frac{1}{\log_3 x - \log_3 3} + \frac{1}{\log_3 x - \log_3 81} = 0\)

    Let \(z = \log_3 x\)

    \(\frac{1}{z}.\frac{1}{z - 1} + \frac{1}{z - 4} = 0\)

    \(z^2 - 4 = 0, z = \pm 2\)

    \(x = 9, \frac{1}{9}\)

75. Given, \(\log_{(2x + 3)}(6x^2 + 23x + 21) = 4 - \log_{(3x + 7)}(4x^2 + 12x + 9)\)

    \(\log_{(2x + 3)}(2x + 3)(3x + 7) = 4 - \log_{(3x + 7)}(2x + 3)^2\)

    \(1 + \log_{(2x + 3)}(3x + 7) = 4 - 2\log_{(3x + 7)}(2x + 3)\)

    Let \(z = \log_{(2x + 3)}(3x + 7)\)

    \(1 + z = 4 - \frac{2}{z}\)

    \(z = 1, 2~\Rightarrow x = -4, -2, \frac{-1}{4}\)

    For logarithm to be defined \(2x + 3 > 0, 2x + 3 \neq 1\) and \(3x + 7> 0, 3x + 7 \neq 1\)

    Thus, \(x = -\frac{1}{4}\) is the only valid solution.

76. Given \(\log_2(x^2 - 1) = \log_{\frac{1}{2}}(x - 1)\)

    \(\log_2(x^2 - 1) = \log_{2^{-1}}(x - 1)\)

    \(\log_2(x^2 - 1) = -\log_2(x - 1) = \log_2(x - 1)^{-1}\)

    \(\log_2(x^2 - 1) = \log_2\frac{1}{x - 1}\)

    \(x^2 - 1 = \frac{1}{x - 1}\)

    \(x=0, x^2 - x - 1 = 0 \Rightarrow x= 0, \frac{1\pm \sqrt{5}}{2}\)

    For logarithm to be defined \(x^2 - 1 > 0\) and \(x - 1 > 0\)

    Thus, \(x = \frac{1 + \sqrt{5}}{2}\) is the only acceptable solution.

77. Given \(\log_5\left(5^{\frac{1}{x} + 125}\right) = \log_5 6 + 1 + \frac{1}{2x}\)

    \(\log_5\left(5^{\frac{1}{x} + 125}\right) - \log_5 6 = 1 + \frac{1}{2x}\)

    \(\log_5\left(\frac{5^{\frac{1}{x} + 125}}{6}\right) = 1 + \frac{1}{2x}\)

    \(\frac{5^{\frac{1}{x} + 125}}{6} = 5^{1 + \frac{1}{2x}}\)

    \(5^{\frac{1}{x}} + 125 = 20.5^{\frac{1}{2x}}\)

    Let \(z = 5^{\frac{1}{x}}\)

    \(z^2 - 30z + 125 = 0\)

    \(z = 5, 25~\Rightarrow x = \frac{1}{2}, \frac{1}{4}\)

78. For \(\log_{100}|x + y| = \frac{1}{2}\)

    \((x + y)^2 = 100\)

    For and \(\log_{10} y - \log_{10}|x| = \log_{100} 4\)

    \(\log_{10}\frac{y}{|x|} = \log_{10}2\)

    \(y = 2|x|~\Rightarrow y^2 = 4x^2\)

    Thus, we get \(5x^2 + 4x|x| = 100\)

    When \(x > 0, x = \frac{10}{3}\)

    When \(x < 0, x = -10\)

    \(y = \frac{20}{3}, 20\)

79. Given, \(2\log_2\log_2 x + \log_{\frac{1}{2}}\log_2(2\sqrt{2}x) = 1\)

    \(\log_2(\log_2 x)^2 - \log_2\log_2(2\sqrt{2}x) = 1\)

    \(\log_2 \left(\frac{(\log_2 x)^2}{\log_2 (2\sqrt{2}x)}\right) = 1\)

    \(\frac{(\log_2 x)^2}{\log_2 (2\sqrt{2}x} = 2\)

    \((\log_ x)^2 = = \log_2(2\sqrt{2}x)\)

    \((\log_2 x)^2 - 3 - 2\log_2 x = 0\)

    \(z^2 -2z - 3 = 0\) where \(z = \log_2\)

    \(z = -1, 3\)

    \(\Rightarrow x = \frac{1}{2}, 8\)

    For log to be defined \(x > 0\) and \(2\sqrt{2}x > 0\)

    \(\log_2 x > 0, \log_2 2\sqrt{2}x > 0\)

    Thus, \(x = 8\) is only acceptable solution.

80. Given \(\log_{\frac{3}{4}}\log_8(x^2 + 7) + \log_{\frac{1}{2}}\log_{\frac{1}{4}}(x^2 + 7)^{-1} = -2\)

    \(\Rightarrow \log_{\frac{3}{4}}\log_{2^3}(x^2 + 7) + \log_{\frac{1}{2}}\log_{2^{-4}}(x^2 + 7)^{-1} = -2\)

    \(\Rightarrow \log_{\frac{3}{4}}\left[\frac{1}{3}\log_2(x^2 + 7)\right] + \log_{\frac{1}{2}}\left[\frac{1}{2}\log_2(x^2 + 7)\right] = -2\)

    Let \(y = \log_2(x^2 + 7),\) then we have

    \(\log_{\frac{3}{4}}\left(\frac{y}{3}\right) + \log_{\frac{1}{2}}\frac{1}{2} + \log_{\frac{1}{}}y = -2\)

    \(\Rightarrow \log_{\frac{3}{4}}y - \log_{\frac{3}{4}}3 + 1+ \log_{2^{-1}}y = -2\)

    \(\Rightarrow \log_2y\left(\log_{\frac{3}{4}}2 - 1\right) = -3 + \log_{\frac{3}{4}} 3\)

    Solving this we get \(\log_2 y = 2 \Rightarrow y = 4\)

    \(x = \pm 3,\) both of these values are valid for logarithm.

81. Given, \(\log_{10}x + \log_{10}x^{\frac{1}{2}} + \log_{10}x^{\frac{1}{4}} \ldots\) to \(\infty = y\)

    \(\log_{10}x\left[1 + \frac{1}{2} + \frac{1}{4}~\text{to}~\infty\right] = y\)

    \(\log_{10}x \frac{1}{1 - \frac{1}{2}} = y \Rightarrow \log_{10}x = \frac{y}{2}\)

    Also, given that

    \(\frac{1 + 3 + 5 + \ldots + (2y - 1)}{4 + 7 + 10 + \ldots + (3y + 1)} = \frac{20}{7\log_10 x}\)

    \(\Rightarrow \frac{\frac{y}{2}[2 + (y - 1)2]}{\frac{y}{2}[8 + (y - 1)3]} = \frac{20}{7\log_{10} x}\)

    \(\frac{2y}{3y + 5} = \frac{20}{7\log_{10} x}\)

    Thus, \(\frac{2y}{3y + 5} = \frac{20\times 2}{7y}\)

    \(\Rightarrow 7y^2 - 60y -100 = 0\)

    \(y = 10, \frac{-10}{7}\)

    Since number of terms cannot be a fraction, therefore \(y = 10\) is the answer. Hence, \(x = 10^5\)

82. Given, \(18^{4x - 3} = (54\sqrt{2})^{3x - 4}\)

    Taking log on both sides,

    \((4x - 3)\log 18 = (3x - 4)log(18.3\sqrt{2})\)

    \(\Rightarrow (4x - 3)\log 18 = (3x - 4)\log 18^{\frac{3}{2}}\)

    \(\Rightarrow 4x - 3 = (3x - 4)\frac{3}{2}\)

    \(\Rightarrow x = 6\)

83. Given, \(4^{\log_9 3} + 9^{\log_2 4} = 10^{\log_x 83}\)

    \(\Rightarrow 4^{\log_{3^2} 3 + 9^{\log_2 2^2}} = 10^{\log_x 83}\)

    \(\Rightarrow 4^{\frac{1}{2}\log_3 3} + 9^{2\log_2 2} = 10^{\log_x 83}\)

    \(\Rightarrow 4^\frac{1}{2} + 9^2 = 10^{\log_x 83}\)

    \(\Rightarrow 83 = 10^{\log_x 83}\)

    \(x = 10\)

84. Given, \(3^{4\log_9 (x + 1)} = 2^{2\log_2 (x + 3)}\)

    \(\Rightarrow 3^{2\log_3 (x + 1)} = x^2 + 3 [\because a^{\log_a N} = N]\)

    \(\Rightarrow 3^{\log_3 (x + 1)^2} = x^2 + 3\)

    \(\Rightarrow x^2 + 2x + 1 = x^2 + 3\)

    \(\Rightarrow x = 1\)

85. Given, \(\frac{6}{5}a^{\log_a x\log_{10} a \log_a 5} - 3^{\log_{10}\left(\frac{x}{10}\right)} = 9^{\log_{100}x + \log_4 2}\)

    \(\Rightarrow \frac{6}{5}a^{\log_{10}x\log_a 5}- 3^{\log_{10}{x - 1}} = 9^{\frac{1}{2}\log_{10} x + \frac{1}{2}\log_2 2}\)

    \(\Rightarrow \frac{6}{5}\left(a^{\log_a 5}\right)^{\log_{10} x} - 3^{\log_{10}{x - 1}} = 3\log_{10}^{x + 1}\)

    \(\Rightarrow \frac{6}{5}5^{\log_{10}x} = 3^{\log_{10}{x + 1}} + 3^{\log_{10}{x + 1}}\)

    \(\Rightarrow 6.5^{\log_{10}{x - 1}} = 3^{\log_{10}{x - 1}}.(1 + 3^2)\)

    \(\Rightarrow \left(\frac{5}{3}\right)^{\log_{10}{x - 1}} = \frac{10}{6}\)

    \(\Rightarrow \log_{10} x - 1 = 1\Rightarrow x = 100\)

86. Given, \(2^{3x + \frac{1}{2}} + 2^{x + \frac{1}{2}} = 2^{\log_2 6}\)

    \(2^{3x}\sqrt{2} + 2^x\sqrt{2} = 6\)

    \((2^x)^3 + 2^x = 3\sqrt{2}\)

    Let \(z = 2^x,\) then we can rewrite above as

    \(z^3 + z = 3\sqrt{2}\)

    \(z = \sqrt{2}, \frac{-\sqrt{2}\pm \sqrt{2 - 12}}{2}\)

    Ignoring complex roots we have \(z = \sqrt{2}\)

    \(\therefore 2^x = \sqrt{2}\)

    \(x = \frac{1}{2}\)

87. Given, \((5 + 2\sqrt{6})^{x^2 - 3} + (5 - 2\sqrt{6})^{x^2 - 3} = 10\)

    \(\Rightarrow (5 + 2\sqrt{6})^{x^2 - 3} + (5 + 2\sqrt{6})^{-(x^2 - 3)} = 10\)

    Let \(z = (5 + 2\sqrt{6})^{x^2 - 3},\) then we can rewrite above as

    \(z + \frac{1}{z} = 10\)

    \(z = 5\pm 2\sqrt{6}\)

    \(\therefore x = \pm 2, \pm\sqrt{2}\)

88. \(2\log_{10}x - \log_x .01 = 2\log_{10}x - \log_x 10^{-2}\)

    \(= 2\log_{10}x + 2\log_x 10 = 2\log_{10}x + 2\frac{1}{\log_{10}x}\)

    \(= 2\left(\log_{10}x + \frac{1}{\log_{10}x}\right)\)

    Let \(z = \log_{10}x,\) then above can be rewritten as

    \(= 2\left(z + \frac{1}{z}\right)\)

    \(= 2\left[\left(\sqrt{z} - \frac{1}{\sqrt{z}}\right)^2 + 2\right] \geq 4\)

    Thus, given condition is proved.

89. Let \(E = \log_b a + \log_a b = \log_b a + \log_a b\)

    Also, let \(z = \log_b a,\) then we can rewrite above as

    \(E = z + \frac{1}{z}\)

    Clearly, \(z \neq 0,\) or the problem will be undefined.

    When \(z > 0, E = z + \frac{1}{z} = \left(\sqrt{z} - \frac{1}{\sqrt{z}}\right)^2 + 2 > 2\)

    When \(z < 0,\) let \(z = -y,\) then we have

    \(E = \left|z + \frac{1}{z}\right| = \left|-y - \frac{1}{y}\right|\)

    \(= y + \frac{1}{y} > 2\)

90. Given, \(\log_{0.3}(x ^2 + 8) > \log_{0.3}9x\)

    \(x^2 + 8 < 9x\)

    \(\Rightarrow 1 < x < 8\)

91. Given, \(\log_{x - 2}(2x - 3) > \log(x - 2)(24 - 6x)\)

    Case I: When \(0 < x - 2 < 1, \Rightarrow 2 < x < 3\)

    Given inequality becomes \(2x - 3 < 24 - 6x\Rightarrow x < \frac{27}{8}\)

    But \(x < 3\) so \(3\) is still limiting value of \(x\)

    Case II: When \(x - 2 > 1, \Rightarrow x > 3\)

    Given inequality becomes \(2x - 3 > 24 - 6x\Rightarrow x > \frac{27}{8}\)

    However, for logarithm function to be defined \(2x - 3 > 0\) and \(24 - 6x > 0\) and also \(x - 2 > 0\)

    Combining all these we get \(2 < x < 3\)

92. Given \(\log_{0.3}(x - 1) < \log_{0.09}(x - 1)\)

    \(\Rightarrow (x - 1)^2 > (x - 1)\)

    \(\Rightarrow x^2 - 3x + 2 > 0\)

    \(\Rightarrow x < 1, x> 2\)

    For logarithm functiton to be defined \(x > 1,\) thus the interval for \(x\) would be \((2, \infty]\)

93. Given, \(\log_{\frac{1}{2}}x \geq \log_{\frac{1}{3}}x\)

    \(\Rightarrow \log_{\frac{1}{2}}x \geq \log_{\frac{1}{2}}x \log_{\frac{1}{3}} \frac{1}{2}\)

    \(\Rightarrow \log_{\frac{1}{2}}x\left[1 - \log_{\frac{1}{3}}\frac{1}{2}\right] \geq 0\)

    \(\Rightarrow \log_{\frac{1}{2}}x\left[1 - \log_{3^{-1}}(2^{-1})\right] \geq 0\)

    \(\Rightarrow \log_{\frac{1}{2}}x[1 - \log_3 2] \geq 0\)

    \(\Rightarrow \log_{\frac{1}{2}}x \geq 0\)

    \(\Rightarrow x \leq 1\)

    For logarithm function to be defined \(x > 0,\) thus range of \(x\) would be \((0, 1]\)

94. Given, \(\log_{\frac{1}{3}}\log_4(x^2 - 5) > 0\)

    \(\log_4(x^2 - 5) < 1\)

    For logarithm ot be defined \(x^2 - 5 > 0\) and \(\log_4 (x^2 - 5) > 0\)

    Combining all these conditions we get two ranges for \(x, (-3, -\sqrt{6})\) and \((\sqrt{6}, 3)\)

95. Given, \(\log (x^2 - 2x - 2)\leq 0\)

    \(\Rightarrow x^2 -2x -2 \leq 1\)

    \(\Rightarrow -1 \leq x \leq 3\)

    For logarithm function to be defined \(x^2 - 2x - 2 > 0\)

    \(x < 1 - \sqrt{3}, x > 1 + \sqrt{3}\)

    Combining these ranges gives us the interval of values in which \(x\) can lie.

96. Given, \(\log_2^2(x - 1)^2 - \log_{0.5}(x - 1) > 5\)

    \(\Rightarrow (2\log_2|x - 1|)^2 - \log_{\frac{1}{2}}(x - 1) > 5\)

    \(\Rightarrow 4[\log_2 (x - 1)]^2 + \log_2(x - 1) > 5\)

    Let \(z = \log_2 (x - 1)\)

    \(\Rightarrow 4z^2 + z - 5 > 0\)

    \(\Rightarrow z < \frac{-5}{4}, x > 1\)

    When \(z < \frac{-5}{4}, x - 1 < 2^{\frac{-5}{4}}\Rightarrow x < 1 + \frac{1}{2\sqrt[4]{2}}\)

    For log to be defined \(x - 1 > 0, x > 1\)

    When \(z > 1, x > 3\)

    Thus range of \(x\) is \(\left(1, 1 + \frac{1}{2\sqrt[4]{2}}\right)\cup (3, \infty)\)

97. Let \(E = \log_2 17\log{\frac{1}{5}} 2\log_3\frac{1}{5} > 2\)

    \(L.H.S. = \log_2 17\log_3 2 = \log_3 17\)

    \(\because 17 > 3^2\)

    \(\therefore \log_3 17 > 2\)

98. We have to prove that \(\frac{1}{3} < \log_{20} 3 < \frac{1}{2}\)

    \(\frac{1}{3} < \log_{20} 3\Rightarrow 1< 3\log_{20}3\)

    \(\Rightarrow 1 < \log_{20} 3^3\Rightarrow 1 < \log_{20} 27\)

    which is true as base is greater than \(1\) and number is greater than base.

    \(\log_{20} 3 < \frac{1}{2}\Rightarrow 2\log_{20} 3 < 1\)

    \(\Rightarrow \log_{20} 3^2 < 1\Rightarrow \log_{20} 9 < 1\)

    which is true because number is less than base.

99. We have to prove that \(\frac{1}{4} < \log_{10} 2 < \frac{1}{2}\)

    Taking first two parts, \(\frac{1}{4} < \log_{10}2\)

    \(\Rightarrow 1 < 4\log_{10} 2 = \log_{10} 2^4 = \log_{10}16\)

    which is true because \(10 > 1\) and number is greater than base.

    Taking last two parts, \(\log_{10} 2 < \frac{1}{2}\)

    \(\Rightarrow 2\log_{10} 2 < 1\Rightarrow \log_{10}2^2 < 1\)

    \(\Rightarrow \log_{10} 4 < 1\)

    which is true because \(10 > 1\) and number is samllers than base.

100. Given, \(\log_{0.1}(4x^2 - 1) > \log_{0.1}3x\)

     \(4x^2 - 3x - 1 < 0\)

     \((4x + 1)(x - 1) < 0\)

     Thus, \([-\infty, -\frac{1}{4})\cup(1, \infty]\) is initial solution.

     Now, \(x > 0\) is another restriction from R.H.S.

     From L.H.S. \(4x^2 - 1 > 0\)

     \(\Rightarrow x < -\frac{1}{2}, x > \frac{1}{2}\)

     Combining all these we get, \(\frac{1}{2} < x < 1\)