15. Logarithm Solutions Part 3

  1. Given, \(\log_2(x^2 - 24) > \log_2 5x\)

    Since base is greater than one, we can say that

    \(x^2 - 24 > 5x\)

    \((x - 8)(x + 3) > 0\)

    \(-3 > x, x > 8\)

    But \(x^2 - 24 > 0\) and also \(x > 0\) for logarithm function to be defined.

    \(\therefore\) the solutions is \(x > 8\)

  2. We have to prove that \(\frac{1}{\log_3\pi} + \frac{1}{\log_4\pi} > 2\)

    \(\Rightarrow \log_{\pi} 3 + \log_{\pi} 4 > 2\)

    \(\Rightarrow \log_{\pi} 12 > 2\)

    \(\Rightarrow 12 > \pi^2\) which is true.

  3. Given \((0.01)^{\frac{1}{3}}\) and \((0.001)^{\frac{1}{5}}\)

    Taking logarithmg of both with base \(10\)

    \(\frac{1}{3}\log_{10} 0.01\) and \(\frac{1}{5}\log_{10}0.001\)

    \(\frac{-2}{3}\) and \(\frac{-3}{5}\)

    Since \(\frac{-3}{5}\) is greater, \(\therefore (0.001)^{\frac{1}{5}}\) is greater.

  4. \(\log_311>\log_39=\log_3(3^2)=2\) and \(\log_23<\log_24=\log_2(2^2)=2\)

    Thus, \(\log_311\) is greater.

  1. \(\log_3(x^2 + 10) > \log_3 7x\)

    \(\Rightarrow x^2 + 10 > 7x\Rightarrow x < 2, x > 5\)

    However, \(x^2 + 10 > 0\) and \(x > 0\)

    Thus, intervals are \(0 < x < 2\) and \(x > 5\)

  1. We have, \(x^{\log_{10} x} > 10\)

    \(\Rightarrow\log_{10}x \log_{10} x > 1\)

    \(\log_{10}x > \pm 1\)

    Thus the range for values of \(x\) would be \((0, 0.1)\cup(10, \infty]\)

  2. We have \(\log_2 x\log_{2x} 2\log_2 4x > 1\)

    \(\frac{1}{\log_x 2}\left[\frac{1}{\log_2 2x}\log_2 2^2 x\right] > 1\)

    \(\frac{1}{\log_x 2}\left[\frac{1}{1 + \log x}\right]\left[2 + \frac{1}{\log_x 2}\right] > 1\)

    Let \(z = \log_x 2\) then the above inequality becomes

    \(\frac{1}{z}\left[\frac{1}{1 + z}\right]\left[2 + \frac{1}{t}\right] > 1\)

    Solving this inequality and applying rules for base of a logarithm we will have following range for \(x, \left(2^{-\sqrt{2}}, \frac{1}{2}\right)\cup\left(1, 2^{\sqrt{2}}\right)\)

  3. Given, \(\log_2 x\log_3 2x + \log_3 x\log_2 4x > 0\)

    Exchaning base, we have \(\log_3 x\log_2 2x + \log_3 x\log_2 4x > 0\)

    \(\log_3 x(\log_2 2 + \log_2 x + \log_2 4 + \log_2 x) > 0\)

    \(\log_3 x(3 + 2\log_2 x) > 0\)

    For, \(\log_3 x > 0, x > 1\)

    and for, \(3 + \log_2 x^2 > 0\)

    \(\log_2 x^2 > -3\)

    Also, for \(\log_3 x < 0, 0 < x < 1\) and for

    \(3 + \log_2 x^2 < 0\)

    \(\log_2 x^2 < -3\)

  4. \(\log_{12}60 = \frac{\log_260}{\log_212} = \frac{\log_2(2^2 \times 3\times 5)}{\log_2 (2^2\times 3)}\)

    \(= \frac{2 + \log_23 + \log_25}{2 + \log_23}\)

    Let \(\log_23 = x\) and \(\log_25 = y\)

    \(\log_{12}60 = \frac{2 + x + y}{2 + x}\)

    Given, \(a = \log_630 = \frac{\log_230}{\log 26} = \frac{\log_2(2\times 3\times 5)}{\log_2(2\times 3)}\)

    \(= \frac{1 + \log_23 + \log_25}{1 + \log_23} = \frac{1 + x + y}{1 + x}\)

    Also given that, \(b= \log_{15}24 = \frac{\log_224}{\log_215} = \frac{\log_2(2^3\times 3)}{\log_2 (3\times 5)}\)

    \(= \frac{3 + \log_23}{\log_23 + \log_25} = \frac{3 + x}{x + y}\)

    From these values of \(a\) and \(b\) in terms of \(x\) and \(y,\) we get

    \(x = \frac{b + 3 - ab}{ab - 3}, y = \frac{2a - b - 2 + ab}{ab - 1}\)

    Substituting these value of \(x\) and \(y\) for \(\log_{12}60,\) we get

    \(\log_{12}60 = \frac{2ab + 2a - 1}{ab + b + 1}\)

  5. Since \(\log_ax, \log_bx\) and \(\log_cx\) are in A.P.

    \(2\log_bx = \log_ax + \log_cx\)

    \(\frac{2}{\log_xb} = \frac{1}{\log_xa} + \frac{1}{\log_xc}\)

    \(\frac{2}{\log_xb} = \frac{\log_xa + \log_xc}{\log_xa\log_xc}\)

    \(\frac{2}{\log_xb} = \frac{\log_xac}{\log_xa\log_xc}\)

    \(2\log_xc = \log_xac\frac{\log_xb}{\log_xa}\)

    \(\log_xc^2 = \log_xac\log_ab\)

    \(c^2 = ac^{\log_ab}\)

  6. We have to use following rules:

    If \(b > 0, N > 0\) or \(b < 0, N< 0,\) then \(\log_bN > 0\)

    else if \(b > 0, N < 0\) or \(b < 0, N > 0,\) then \(\log_bN < 0\)

    \(a = \log_{\frac{1}{2}}(\sqrt{0.125}) > 0\) because both base and number are less than \(0\)

    \(b = \log_3\left(\frac{1}{\sqrt{24} - \sqrt{17}}\right)\)

    \(= \log_3\left(\frac{\sqrt{24} + \sqrt{17}}{(\sqrt{24} - \sqrt{17)(\sqrt{14} + \sqrt{17})}}\right)\)

    \(= \log_3 \left(\frac{\sqrt{24} + \sqrt{17}}{7}\right) > 0\) because both base and number are greater than \(0\)

  7. Given, \(e^{\frac{-\pi}{2}} < \theta < \frac{\pi}{2}\)

    Taking log natural of both sides

    \(\log_ee^{-\frac{\pi}{2}} < \log_e\theta < \log_e\frac{\pi}{2}\)

    \(\Rightarrow -\frac{\pi}{2} < \log_e\theta < 1< \frac{\pi}{2}\left[\because \log_e \frac{\pi}{2} < \log_ee\right]\)

    \(\Rightarrow -\frac{\pi}{2} < \log_e\theta < \frac{\pi}{2}\)

    \(\Rightarrow \cos(\log_e\theta) > 0\)

    Also, \(e^{\frac{-\pi}{2}} < \theta < \frac{\pi}{2}\)

    \(\Rightarrow 0 < \theta < \frac{\pi}{2}\left[\because e^{-\frac{\pi}{2}} > 0\right]\)

    \(\Rightarrow 0 < \cos\theta < 1\)

    \(\Rightarrow \log_e \cos\theta < 0\)

    \(\therefore \cos(\log_e\theta > \log_e(\cos\theta))\)

  8. Given, \(\log_2 x + \log_2 y \geq 6\)

    \(\log_2xy \geq 6,\Rightarrow xy \geq 64\)

    This means \(x\) and \(y\) are positive as negative vlaues will not be valid for logarithm function.

    A.M. \(\geq\) G.M.

    \(\frac{x + y}{2} \geq \sqrt{xy}\)

    \(x + y \geq 16\)

  9. Given, \(\log_b a \log_c a - \log_a a + \log_a b\log_c b - \log_b b + \log_a c \log _b c - \log_c c = 0\)

    \(\frac{(\log_a)^2}{\log b \log c} - 1 + \frac{(\log b)^2}{\log a \log c} - 1 + \frac{(\log c)^2}{\log a\log b} - 1 = 0\)

    Let \(\log a = x, \log b = y, \log c = z,\) then we have

    \(\frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy} - 3 = 0\)

    \(\frac{x^3 + y^3 + z^3 - 2xyz}{xyz} = 0\)

    \((x + y + z)(x^2 + y^2 + z^2 - xy - yz - xz) = 0\)

    \((x + y + z)\frac{1}{2}[(x - y)^2 + (y - z)^2 + (z - x)^2] = 0\)

    \(\because x, y , z\) are different the term inside brackets will be always positive. Thus,

    \(x + y + z = 0,\) now substituting the original values we get

    \(\log abc = 0, \Rightarrow abc = 1\)

  10. Given, \(y = 10^{\frac{1}{1 - \log x}}\)

    Taking logarithm of both sides we get

    \(\log y = \frac{1}{1 - \log x}\)

    Also, \(z = 10^{\frac{1}{1 - \log y}}\)

    \(\log z = \frac{1}{1 - \log y}\)

    \(\Rightarrow log y = 1 - \frac{1}{\log z}\)

    Equating the values for \(\log y,\) we get

    \(\frac{1}{1 - \log x} = 1 - \frac{1}{\log z}\)

    \(\log x = \frac{1}{1 - \log z}\)

    \(x = 10^{\frac{1}{1 - \log z}}\)

  11. Since \(n\) is natural numbers and \(p_1, p_2, p_3, \ldots, p_k\) are distinct primes, therefore \(a_1, a_2, \ldots, a_k\) are also natural numbers.

    Now, \(n = p_1^{a_1}p_2^{a_2}p_3^{a_3} \ldots p_k^{a_k}\)

    \(\log n = a_1\log p_1 + a_2\log p_2 + \ldots + a_k \log p_k\)

    \(\log n \geq \log 2 + \log 2 + \ldots + \log 2\) [since bases are primes so minimum value if \(2\) and powers are natural numbers so they are greater than 1]

    \(\log n \geq k\log 2\)

  12. Let \(d\) be the common difference of this A.P. then we can write

    \(3\log_y x = 3 +d\Rightarrow \log_yx^3 = 3 + d\Rightarrow x^3 = y^{(3 + d)}\)

    \(3\log_z y = 3 + 2d\Rightarrow y^3 = z^{(3 + 2d)}\)

    \(7\log_x z = 3 + 3d\Rightarrow z^7 = x^{(3 + 3d)}\)

    \(y^3 = z^{(3 + 2d)}\Rightarrow y = z^{\frac{3 + 2d}{3}}\)

    \(x^3 = y^{(3 + d)}\Rightarrow x = y^{\frac{3 + d}{3}} = z^{\frac{(3 + d)(3 + 2d)}{9}}\)

    \(z^7 = x^{(3 + 3d)}\Rightarrow x = z^\frac{7}{3 + 3d}\)

    Thus, \(\frac{(3 + d)(3 + 2d)}{9} = \frac{7}{3 + 3d}\)

    \(\Rightarrow d = \frac{1}{2}\)

    Thus, \(x^{18} = y^{21} = z^{28}\)

  13. We have \(\log_418 = \log_{2^2}(2\times 3^2) = \frac{1}{2} + \log_23\)

    Thus, it will be enough to prove that \(\log_23\) is an irrational number.

    Let \(\log_23 = \frac{p}{q}\) where \(p, q \in I\)

    \(2^{\frac{p}{q}} = 3\Rightarrow 2^p = 3^q\)

    However, \(2^p\) is an even number while \(3^q\) is an odd number, and hence the equality will never be achieved. Therefore \(\log_23\) is an irrational number.

  14. \(\because x, y, z\) are in G.P.

    \(\frac{y}{x} = \frac{z}{y}\)

    \(\ln \frac{y}{x} = \ln \frac{z}{y}\)

    \(\ln y - \ln x = \ln z - \ln x\)

    \(\ln x, \ln y, \ln z\) are in A.P.

    \(1 + \ln x, 1 + \ln y, 1 + \ln z\) are in A.P.

    \(\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}\) are in H.P.

  15. \(\log_{30}8 = \log_{30}2^3 = 3\log_{30}2 = 3\log_{30}\frac{30}{15}\)

    \(= 3\log_{30}30 - 3\log_{30}15 = 3 - 3(\log_{30}3 + \log_{30}5)\)

    \(= 3(1 - a - b)\)

  16. Given, \(log_7 12 = a\) and \(\log_{12} 24 = b\)

    Multiplying \(ab = \log_724\)

    Adding \(1\) on both sides

    \(ab + 1 = \log_724 + \log_77 = \log_7168\)

    Similarly, \(8a = \log_712^8\) and \(5ab = \log_7168^5\)

    \(\frac{ab + 1}{8a - 5ab} = \frac{\log_7168}{\log_712^8 - \log_7168^5}\)

    Upong simplification we find that \(\log_{54}168 = \frac{ab + 1}{8a - 5ab}\)

  17. Case I: When \(x > 1,\) \(x > a^1 + 1.\) Also, \(a^ + 1 < 1 \therefore x > 1\)

    Case II: When \(x < 1, x < a^2 + 1.\) Also, \(a^2 > 0, \therefore x < 1\)

    In both the cases \(x > 0\)

  18. Given \(\log_{12}18 = a\) and \(\log_{24}54 = b\)

    \(ab + 5(a - b) = \frac{\log 18}{\log 12}\frac{\log 54}{\log 24} + 5\left(\frac{\log 18}{\log 12} - \frac{\log 54}{\log 24}\right)\)

    \(=\frac{\log 18\log 54 + 5(\log 18\log 24 - \log 54\log 12)}{\log 12 \log 24}\)

    \(\log 18 = \log 2 + 2\log 3, \log 12 = 2\log 2 + \log 3, \\\log 24 = 3\log 2 + \log 3, \log 54 = \log 2 + 3\log 3\)

    Substituting and simplifying we obtain the desired result.

  19. Since \(a, b, c\) are in G.P.

    \(\Rightarrow b^2 = ac\)

    Taking logarithm with base \(x,\) we get

    \(2\log_x b = \log_x a + \log_x c\)

    \(\frac{2}{\log_b x} = \frac{1}{\log_a x} + \frac{1}{\log_c x}\)

    Thus, \(\log_a x, \log_b x, \log_c x\) are in H.P.

  20. Let \(r\) be the common ratio of G.P. and \(d\) be the common difference.

    \(\log a_n - b_n = \log a + n\log r - (b + nd) = \log a - b\)

    \(n\log r - nd = 0\)

    \(\log r = d\)

    Thus, base \(b = r^\frac{1}{d}\)

  21. Given, \(\log_3 2, \log_3(2^x - 5)\) and \(\log_3\left(2^x - \frac{7}{2}\right)\) are in A.P.

    \(\therefore 2\log_3(2^x - 5) = \log_3\left(2^x - \frac{7}{2}\right) + \log_3 2\)

    \(\Rightarrow (2^x - 5)^2 = 2\left(2^x - \frac{7}{2}\right)\)

    Let \(z = 2^x,\) then we have

    \(z^2 - 10z + 25 = 2z - 7\)

    \(z^2 -12z + 32 = 0\)

    \(z = \frac{12 \pm \sqrt{144 - 128}}{2} = 8, 4\)

    \(\Rightarrow x = 2, 3\)

    But if \(x = 2, 2^x - 5 < 0\) so only acceptable value is \(x = 3\)

  22. Let \(\log_2 7\) is a rational number i.e. \(\log_2 7 = \frac{p}{q}\) where \(p, q \in I\)

    \(7 = 2^{\frac{p}{q}}\Rightarrow 7^q = 2^p\)

    However, integral power of \(7\) is odd but integral power of \(2\) is even. Thus equality cannot hold and \(\log_2 7\) cannot be rational i.e. it is irrational number.

  23. Given, \(\log_{0.5}(x - 2) < \log_{0.25}(x - 2)\)

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

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

    Thus, \(x > 3\) for logarithm function to be defined.