# 13. Logarithm Solutions Part 2

1. 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 $$28$$

2. 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 $$5^3 - 5^2 = 100$$

3. 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 $$11$$

4. 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 $$16$$

5. 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 $$18$$

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

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

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

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

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

$$1 + \log_p x = 1$$

$$\log_p x = 0$$

$$x = 1$$

7. 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

$$x + 5 = 25 + x -10\sqrt{x}$$

$$\sqrt{x} = 2~\Rightarrow x = 4$$

8. $$\log_2 x + \log_4 (x + 2) = 2$$

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

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

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

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

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

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

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

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

$$\therefore \frac{1}{z} + z = \frac{5}{2}$$

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

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

$$x + 2 = 2^x, x + 2 = \sqrt{x}$$

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

$$\therefore x = 2$$

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

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

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

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

11. $$2\log_x a + \log_{ax} a + 3\log_{a^2x} a = 0$$

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

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

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

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

$$\frac{2}{z} + \frac{1}{z + 1} + \frac{3}{z + 2} = 0$$

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

$$2z^2 + 6z + 4 + z^2 + 2z + 3z^2 + 3z = 0$$

$$6z^2 + 11z + 4 = 0$$

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

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

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

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

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

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

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

$$2^x = 2, 1 + 2^x = 3~\Rightarrow x = 1$$

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

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 $$x = 2, \frac{1}{4}, \frac{1}{\sqrt[3]{2}}$$

14. 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 $$\log_3x \neq 0$$

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

15. 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.

16. 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 $$z = \log_8 x$$

$$1 - 2z = 3z^2$$

$$z = -1, \frac{1}{3}$$

$$x = 2, \frac{1}{8}$$

17. 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 $$x = 2$$

18. 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 $$z = \log_{10}x$$

$$2z + \frac{2}{z} = 5$$

$$2z^2 - 5z + 2 = 0$$

$$z = 2, \frac{1}{2}$$

$$x = 100, \sqrt{10}$$

19. 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 $$e$$:

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

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

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

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

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

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

20. 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 $$x = 2$$

21. 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.

22. 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 $$x = -4$$

23. 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

$$0.30103x + 0.95424x = 2$$

$$x = 10593$$

24. 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 $$z = \log_3 x$$

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

$$z^2 - 4 = 0, z = \pm 2$$

$$x = 9, \frac{1}{9}$$

25. 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.

26. 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.

27. 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 $$z = 5^{\frac{1}{x}}$$

$$z^2 - 30z + 125 = 0$$

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

28. 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 $$x < 0, x = -10$$

$$y = \frac{20}{3}, 20$$

29. 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.

30. 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.

31. 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, $$x = 10^5$$

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

Taking log on both sides,

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

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

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

$$\Rightarrow x = 6$$

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

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

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

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

$$\Rightarrow 83 = 10^{\log_x 83}$$

$$x = 10$$

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

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

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

$$\Rightarrow x^2 + 2x + 1 = x^2 + 3$$

$$\Rightarrow x = 1$$

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

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

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

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

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

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

$$\Rightarrow \log_{10} x - 1 = 1\Rightarrow x = 100$$

36. 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 $$z = \sqrt{2}$$

$$\therefore 2^x = \sqrt{2}$$

$$x = \frac{1}{2}$$

37. 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

$$z + \frac{1}{z} = 10$$

$$z = 5\pm 2\sqrt{6}$$

$$\therefore x = \pm 2, \pm\sqrt{2}$$

38. $$2\log_{10}x - \log_x .01 = 2\log_{10}x - \log_x 10^{-2}$$

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

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

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.

39. 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

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

$$= y + \frac{1}{y} > 2$$

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

$$x^2 + 8 < 9x$$

$$\Rightarrow 1 < x < 8$$

41. 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 $$2 < x < 3$$

42. 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 $$(2, \infty]$$

43. 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 $$(0, 1]$$

44. 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 $$(\sqrt{6}, 3)$$

45. 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.

46. 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 $$\left(1, 1 + \frac{1}{2\sqrt[4]{2}}\right)\cup (3, \infty)$$

47. Let $$E = \log_2 17\log{\frac{1}{5}} 2\log_3\frac{1}{5} > 2$$

$$L.H.S. = \log_2 17\log_3 2 = \log_3 17$$

$$\because 17 > 3^2$$

$$\therefore \log_3 17 > 2$$

48. 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.

49. 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.

50. 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, $$\frac{1}{2} < x < 1$$