# 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.