# 11. Logarithm Solutions Part 1

1. $$\log_{\sqrt{8}}x = \frac{10}{3} \Rightarrow \log_2^{\frac{3}{2}} x = \frac{10}{3}$$

$$\Rightarrow \frac{2}{3}\log_2 x = \frac{10}{3} \Rightarrow \log_2 x = 5$$

$$\Rightarrow x = 25$$

2. L.H.S. $$= \log_b a.\log_c b.\log_a c = \frac{\log a}{\log b}\frac{\log b}{\log c}\frac{\log c}{\log a} = 1$$

3. L.H.S. $$= \log_3 \log_2 \log_{\sqrt{5}} (625) = \log_3 \log_2 \log_{\sqrt{5}} 5^4$$

$$= \log_3 \log_2 8 = \log_3 \log_2 2^3 = \log_3 3 = 1$$

4. Given, $$a^2 + b^2 = 23ab \Rightarrow (a + b)^2 = 25ab$$

$$\Rightarrow \left(\frac{a + b}{5}\right) = \sqrt{ab}$$

Taking log on both sides, we get

$$\log \frac{a + b}{5} = \frac{1}{2}(\log a + \log b)$$

5. L.H.S. $$= 7\log \frac{16}{15} + 5\log \frac{25}{24} + 3\log \frac{81}{80}$$

$$= 7[\log 2^4 - \log(3.5)] + 5[\log 5^2 - \log 2.2.2.3] + 3[\log 3^4 - \log 2^4.5]$$

$$= 7[4\log2 - \log 3 - \log 5] + 5[2\log 5 - 3\log 2 -\log 3] + 3[4\log 3 - 4\log 2 -\log 5]$$

$$= \log 2 =$$ R.H.S.

6. L.H.S. $$=\log\tan 1^{\circ} + \log\tan 2^{\circ} + \ldots + \log \tan 89^{\circ}$$

$$= \log(\tan 1^{\circ}.\tan 89^{\circ}).\log(\tan 2^{\circ}.\tan 88^{\circ}) \ldots \log \tan 45^{\circ}$$

$$= \log(\tan 1^{\circ}.\cot 1^{\circ}).\log(\tan 2^{\circ}.\cot 2^{\circ}) \ldots \log \tan 45^{\circ}$$

$$= \log 1.\log 1\ldots \log 1 = 0$$

7. $$\log_9 \tan \frac{\pi}{6} = \log_9 \frac{1}{\sqrt{3}} = \log_3^2 3^{-\frac{1}{2}}$$

$$= \frac{1}{2}\log_3 3^{-\frac{1}{2}} = \frac{1}{2}.-\frac{1}{2} = -\frac{1}{4}$$

8. $$\frac{\log_{a^2}b}{\log_{\sqrt{a}}(b)^2} = \frac{\frac{1}{2}\log_a b}{2.2\log_a b}$$

$$= \frac{\log_a b}{8\log_a b} = \frac{1}{8}$$

9. $$\log_{\sqrt{5}}.008 = 2\log_5 \frac{8}{1000} = 2[\log_5 8 - \log_5 1000]$$

$$= 2[\log_5 8 - \log_5 8 - \log 125] = -6$$

10. $$\log_{2\sqrt{3}}144 = \log_{2\sqrt{3}}12^2 = \log_{2\sqrt{3}}(2\sqrt{3})^4 = 4$$

11. L.H.S. $$=\log_3 \log_2 \log_{\sqrt{3}} 81 = \log_3 \log_2 \log_{\sqrt{3}}(\sqrt{3})^8$$

$$= \log_3 \log_2 8 = \log_3 \log_2 2^3 = \log_3 3 = 1$$

12. L.H.S. $$= \log_a x \log_b y = \frac{\log x}{\log a}\frac{\log y}{\log b}$$

$$= \frac{\log x}{\log b}\frac{\log y}{\log a} = \log_b x \log_a y =$$ R.H.S.

13. L.H.S. $$= \log_2 \log_2 \log_2 16 = \log_2 \log_2 \log_2 2^4 = \log_2 \log_2 4 = \log_2 \log_2 2^2 = \log_2 2 = 1 =$$ R.H.S.

14. R.H.S. $$= \log_b x\log_c b \ldots \log_n m \log_a n = \frac{\log x}{\log b}\frac{\log b}{\log c}\ldots \frac{\log m}{\log n}\frac{\log n}{\log a}$$

$$= \frac{\log x}{\log a} = \log_a x =$$ L.H.S.

15. R.H.S. $$= 10^x\log_{10}a = z$$ (say)

Taking log of both sides with base $$10$$

$$\log_{10} z =x\log_{10}a = \log_{10}a^x \Rightarrow z = a^x =$$ L.H.S.

16. Given, $$a^2 + b^2 = 7ab \Rightarrow a^2 + b^2 + 2ab = 9ab$$

$$\Rightarrow \left(\frac{a + b}{3}\right)^2 = ab$$

Taking log of both sides

$$\Rightarrow 2\log \left(\frac{a + b}{3}\right) = \log ab = \log a + \log b$$

$$\Rightarrow \log \left\{\frac{1}{3}(a + b)\right\} = \frac{1}{2}(\log a + \log b)$$

17. L.H.S. $$= \frac{\log a(\log_b a)}{\log b(\log_a b)} = \frac{\log a \frac{\log a}{\log b}}{\log b \frac{\log a}{\log b}}$$

$$= \frac{\log a\log a - \log a \log b}{\log b \log b - \log b \log a}$$

$$\frac{\log a(\log a - \log b)}{\log b(\log b - \log a)} = - \log_b a =$$ R.H.S.

18. L.H.S. $$= \log(1 + 2 + 3) = \log 6 = \log 2.3 = 0 + \log 2 + \log 3$$

$$= \log 1 + \log 2 + \log 3[\because \log 1 = 0]$$

19. L.H.S. $$= 2\log(1 + 2 + 4 + 7 + 14) = 2\log 28 = \log 28^2 = \log 784$$

$$= \log 1*2*4*7*14 = \log 1 + \log 2 + \log 4 + \log 7 + \log 14$$

20. L.H.S. $$= \log 2 + 16\log \frac{16}{15} + 12\log \frac{25}{24} + 7\log \frac{81}{80} = 1$$

$$= \log 2 + 16[\log 16 - \log 15] + 12[\log 25 - \log 24] + 7[\log 81 - \log 80]$$

$$= \log 2 + 16[\log 2^4 - \log 3*5] + 12[\log 5^2 - \log 2^3*3] + 7[\log 3^4 - \log 2^4*5]$$

$$= \log 2 + 16[4\log 2 - \log 3 -\log 5] + 12[2\log 5 - 3\log 2 -\log 3] + 7[4\log 3 - 4\log 2 -\log 5]$$

$$= \log 2[1 + 64 - 36 - 28] + \log 3[28 - 16 - 12] + \log 5[24 - 7 - 16]$$

$$= \log 2 + \log 5 = \log 10 = 1$$ [default base is $$10$$ for common logarithms]

21. Given, $$\frac{\log_9 11}{\log_5 13} \div \frac{\log_3 11}{\log_{\sqrt{5}} 13}$$

$$= \frac{\log_{3^2} 11}{\log_5 13}\frac{\log_{5^{\frac{1}{2}}} 13}{\log_3 11}$$

$$= \frac{\frac{1}{2}\log_3 11}{\log_5 13}\frac{2\log 13}{\log_3 11} = 1$$

22. Given, $$3^{\sqrt{\log_3 2}} - 2^{\sqrt{\log_2 3}}$$

Taking log with base $$10,$$ we get

$$=\sqrt{\log_3 2}\log 3 - \sqrt{\log_2 3}\log 2$$

$$= \sqrt{\frac{\log 2}{\log 3}(\log 3)^2} - \sqrt{\frac{\log 3}{\log 2}(\log 2)^2}$$

$$= \sqrt{\log 2\log 3} - \sqrt{\log 3\log 2} = 0$$

23. Given $$\log_{10} 343 = 2.5353 \Rightarrow \log_{10} 7^3 = 2.5353 \Rightarrow \log_{10} 7 = .8451$$

For, $$7^n > 10^5 \Rightarrow n\log_{10}7 = 5\Rightarrow n = \frac{5}{.8451}$$

Thus, least such integer is $$6$$

24. Since $$a, b, c$$ are in G.P., we can write $$b^2 = ac$$

Taking log of both sides

$$2\log b = \log a + \log c$$

Thus, $$\log a, \log b, \log c$$ are in A.P.

i.e. $$\frac{1}{\log a}, \frac{1}{\log b}, \frac{1}{\log c}$$ are in H.P.

Multiplying each term with $$\log x$$

i.e. $$\frac{\log x}{\log a}, \frac{\log x}{\log b}, \frac{\log x}{\log c}$$ are in H.P.

i.e. $$\log_a x, \log_b x, \log_c x$$ are in H.P.

25. L.H.S. $$= \log \sin 8x = \log 2\sin 4x \cos 4x$$

$$= \log 2 + \log \sin 4x + \log \cos 4x$$

$$= \log 2 + \log 2\sin 2x\cos 2x + \log \cos 4x$$

$$= 2\log 2 + \log \sin 2x + \log \cos 2x + \log \cos 4x$$

$$= 2\log 2 + \log 2\sin x\cos x + \log \cos 2x + \log \cos 4x$$

$$= 3\log 2 + \log \sin x + \log \cos x + \log \cos 2x + \log \cos 4x$$

26. We have to prove $$xyz + 1 = 2yz$$

Dividing both sides by $$yz,$$ we get

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

L.H.S. $$= \log_{2a}a + \frac{1}{\log_{3a}2a\log_{4a}3a}$$

$$= \frac{\log a}{\log 2a} + \frac{\log 3a}{\log 2a}\frac{\log 4a}{\log 3a}$$

$$=\frac{\log a}{\log 2a} + \frac{\log 4a}{\log 2a}$$

$$= \frac{\log a + \log 4a}{\log 2a} = \frac{\log (2a)^2}{\log 2a} = 2$$

27. We have to prove that $$\log_{c + b} a + \log_{c - b} a = 2\log_{c + b}a\log_{c -b}a$$

Dividing both sides by $$\log_{c + b}a\log_{c -b}a,$$ we get

$$\frac{1}{\log_{c - b}a} + \frac{1}{\log_{c + b}a} = 2$$

$$\log_a(c - b) + \log_a(c + b) = 2$$

$$\log_a(c^2 - b^2) = 2$$

$$c^2 - b^2 = a^2$$

$$c^2 = a^2 + b^2$$ which is true because $$c$$ is hypotenuse and $$a$$ and $$b$$ are sided of a right angle triangle.

28. We have to prove that, $$x^xy^yz^z = 1$$

Taking log of both sides

$$x\log x + y\log y + z\log z = 0$$

Let $$\frac{\log x}{y - z} = \frac{\log y}{z - x} = \frac{\log z}{x - y} = k$$

$$\log x = k(y - z), \log y = k(z - x), \log z = k(x - y)$$

$$x\log x + y\log y + z\log z = k(xy - zx + yz - xy + zx - yz) = 0$$

29. Given, $$\frac{yz\log(yz)}{y + z} = \frac{zx\log(zx)}{z + x} = \frac{xy\log(xy)}{x + y}$$

Dividing each term by $$xyz,$$ we get

$$\frac{\log(yz)}{x(y + z)} = \frac{\log(zx)}{y(z + x)} = \frac{\log(xy)}{z(x + y)}$$

$$\frac{\log y + \log z}{xy + zx} = \frac{\log z + \log x}{yz + xy} = \frac{\log x + \log y}{zx + yz} = k$$ (let)

$$\log y + \log z = k(xy + zx), \log z + \log x = k(yz + zx), \log x + \log y = k(zx + yz)$$

Adding all these, we get

$$2(\log x + \log y + \log z) = 2k(xy + yz + xz)$$

$$(\log x + \log y + \log z) = k(xy + yz + xz)$$

$$\therefore \log x = kyz \Rightarrow x\log x = kxyz$$

Similarly $$y\log y = kxyz$$ and $$z\log z = kxyz$$

$$\therefore x\log x = y\log y = z\log z = kxyz$$

$$\Rightarrow x^x = y^y = z^z$$

30. We have to prove that $$(yz)^{\log y - \log z}(zx)^{\log z - \log x}(xy)^{\log x -\log y} = 1$$

Taking log of both sides

$$(\log y - \log z)(\log y + \log z) + (\log z - \log x)(\log z - \log x) + (\log x - \log y)(\log x + \log y) = 0$$

$$(\log y)^2 - (\log z)^2 + (\log z)^2 - (\log x)^2 + (\log x)^2 - (\log y)^2 = 0$$

$$0 = 0$$

31. We have to prove that $$\frac{1}{\log_2 N} + \frac{1}{\log_3 N} + \ldots + \frac{1}{\log_{1988} N} = \frac{1}{\log_{1988!} N}$$

L.H.S. $$= \log_N 2 + \log_N 3 + \ldots + \log_N 1988$$

$$= \log_N (2.3.4\ldots 1988)$$

$$= \log_N 1988! = \frac{1}{\log_{1988!}N}$$

32. L.H.S. $$= \log(1 + x) + \log(1 + x^2) + \log(1 + x^4) \ldots$$ to $$\infty$$

$$\log (1 + x + x^3 + x^4 + \ldots)$$ to $$\infty$$

$$= \log \frac{1}{1 - x}~[\because |x| < 1]$$

$$= -\log(1 - x)$$

33. Let $$S_n = \frac{1}{\log_2 a} + \frac{1}{\log_4 a} + \ldots$$ upto $$n$$ terms`

$$S_n = \log_a 2 + \log_a 4 + \log_a 8 + \ldots$$ upto $$n$$ terms

$$S_n = (1 + 2 + 3 + \ldots + n)\log_a 2$$

$$= \frac{n(n + 1)}{2}\log_a 2$$

34. L.H.S. $$= \frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1}$$

$$= \frac{1}{\log_4 10 + \log_{4} 4} + \frac{1}{\log_{2}20 + \log_2 2} + \frac{1}{\log_5 8 + \log_5 5}$$

$$= \frac{1}{\log_4 40} + \frac{1}{\log_2 40} + \frac{1}{\log_4 40}$$

$$= \log_{40} 4 + \log_{40} 2 + \log_{40} 5 = \log_{40} 40 = 1$$

35. We have to prove that $$\frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} = 1$$

L.H.S. $$= \frac{1}{=log_a{bc} + 1} + \frac{1}{\log_b(ca) + 1} + \frac{1}{\log_c(ab) + 1}$$

$$= \frac{1}{=log_a{bc} + \log_a a} + \frac{1}{\log_b(ca) + \log_b b} + \frac{1}{\log_c(ab) + \log_c c}$$

$$= \frac{1}{\log_a(abc)} + \frac{1}{\log_b(abc)} + \frac{1}{\log_c(abc)}$$

$$= \log_{abc}a + \log_{abc}b + \log_{abc}c = \log{abc}abc = 1$$

36. We have to prove that $$\frac{1}{1 + \log_b a + \log_b c} + \frac{1}{1 + \log_c a + \log_c b} + \frac{1}{1 + \log_a b + \log_a c} = 1$$

L.H.S. $$= \frac{1}{\log_b b + \log_b a + \log_b c} + \frac{1}{\log_c c + \log_c a + \log_c b} + \frac{1}{\log_a a + \log_a b + \log_a c}$$

$$= \frac{1}{\log_b abc} + \frac{1}{\log_c abc} + \frac{1}{\log_a abc}$$

Like previus exercise above expression evaluates to $$1$$

37. We have to prove that $$x^{\log y - \log z}y^{\log z - \log x}z^{\log x - \log y} = 1$$

Taking log of both sides

$$(\log y - \log z)\log x + (\log z - \log x)\log y + (\log x - \log y)\log z = 0$$

$$\log y\log x - \log z\log x + \log z\log y - \log x\log y + \log x\log z - \log y\log z = 0$$

$$0 = 0$$

38. We have to prove that $$a^xb^yc^z = 1$$

$$x\log a + y\log b + z\log c = 0$$

Let $$\frac{\log a}{y - z} = \frac{\log b}{z - x} = \frac{\log c}{x - y} = k$$

$$x\log a = k(xy - zx), y\log b = k(yz - xy), z\log c = k(zx - yz)$$

$$x\log a + y\log b + z\log c = k(xy -zx + yz - xy + zx - yz) = 0$$

39. Let $$\frac{x(y + z - x)}{\log x} = \frac{y(z + x - y)}{\log y} = \frac{z(x + y - z)}{\log z} = \frac{1}{k}$$

$$\log x = kx(y + z - x), \log y = ky(z + x - y), \log z = kz(x + y - z)$$

Let $$y^zz^y = z^xx^z = x^yy^x = c$$

Taking log $$z\log y + y\log z = x\log z + z\log x = y\log x + x\log y = log c$$

$$\Rightarrow zky(z + x - y) + ykz(x + y - z) = xkz(x + y - z) + zkx(y + z - x)\\ = ykx(y + z - x) + xky(x + z - y)$$

$$\Rightarrow yz^2 + xyz - y^2z + xyz + y^2z - z^2y = x^2z + xyz -xz^2 + xyz + xz^2 - x^2z\\ = xy^2 + xyz - x^2y + x^2y + xyz - xy^2$$

$$\Rightarrow 2xyz = 2xyz = 2xyz$$

40. Let $$\frac{\log a}{b - c} = \frac{\log b}{c - a} = \frac{\log c}{a - b} = k$$

$$\log a = k(b - c), \log b = k(c - a), \log c = k(a - b)$$

We have to prove that $$a^{b + c}b^{c + a}c^{a + b} = 1$$

Taking log of both sides, we get

$$(b + c)\log a + (c + a)\log b + (a + b)\log c = 0$$

Substituting the values, we get

$$k(b^2 - c^2) + k(c^2 - a^2) + k(a^2 - b^2) = 0$$

41. Let $$\frac{\log x}{q - r} = \frac{\log y}{r - p} = \frac{\log z}{p - q} = k$$

$$\log x = k(q - r), \log y = k(r - p), \log z = k(p - q)$$

We have to prove that $$x^{q + r}y^{p + r}z^{p + q} = x^py^qz^r$$

Taking log of both sides, we get

$$(q + r)\log x + (p + r)\log y + (p + q)\log z = p\log x + q\log y + r\log z$$

Substituting the value of $$\log x, \log y, \log z$$

$$k(q^2 - r^2) + k(r^2 - p^2) + k(p^2 - q^2) = k(pq - pr + qr - pq + pq - qa)$$

$$0 = 0$$

42. Given $$y = a^{\frac{1}{1 - \log_a x}}$$ and $$z = a^{\frac{1}{1 - \log_a y}}$$

$$z = a^{\frac{1}{1 - \log_a (a^{\frac{1}{1 - \log_a x}})}}$$

$$z = a^{\frac{1}{1 - \frac{1}{1 - \log_a x}}}$$

Taking log of both sides with base $$a,$$ we get

$$\log_a z = \frac{1}{1- \frac{1}{1 - \log_a x}}$$

$$\log_a z = \frac{1- \log_a x}{-\log_a x} = 1 - \frac{1}{\log_a x}$$

$$x = a^{\frac{1}{1 - \log_a z}}$$

43. Given $$f(y) = e^{f(z)}$$ and $$z= e^{f(x)},$$ where $$f(x) = \frac{1}{1 - \log_e x}$$

$$f(y) = e^{\frac{1}{1 - \log_e z}}$$

$$z = e^{\frac{1}{1- \log_e x}}$$

$$f(y) = e^{\frac{1}{1 - \log_e z}} = e^{\frac{1}{1 - \log_e e^{\frac{1}{1 - \log_e x}}}}$$

$$f(y) = e^{\frac{1}{1 - \frac{1}{1 -\log_e x}}}$$

Following like exercise above

$$x = e^{f(y)}$$

44. L.H.S. $$= \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \ldots + \frac{1}{\log_{43} n}$$

$$= \log_n 2 + \log_n 3 + \log_n 4 + \ldots + \log_n 43$$

$$=\log_n(2.3.4\ldots 43) = \log_n 43! = \frac{1}{\log_{43!}n} =$$ R.H.S.

45. L.H.S. $$= 2(\log a + \log a^2 + \log a^3 + \ldots + \log a^n)$$

$$=2\log a(1 + 2 + 3 + \ldots + n) = 2\log a \frac{n(n + 1)}{2}$$

$$= n(n + 1)\log a$$

46. We will make use of the fact that positive characteristics of $$n$$ of a logarithm means there are $$n + 1$$ digits in the number.

Let $$\log y = 12\log 12 = 12\log 2.2.3$$

$$= 12[2*.301 + .477] = 12.96$$

Thus, number of digits is $$13$$.

47. Here we can use the result that number of positive integers having base $$b$$ and characteristics $$n$$ is $$b^{n + 1}- b^n$$

Thus, number of integers with base $$3$$ and characteristics $$2$$ is $$3^3 - 3^2 = 18.$$

48. L.H.S. $$= \log_a x\log_b y$$

$$= \frac{\log x}{\log a}\frac{\log y}{\log b}$$

$$= \frac{\log x}{\log b}\frac{\log y}{\log a}$$

$$\log_bx\log_ay =$$ R.H.S.

49. Given, $$a, b, c$$ are in G.P.

$$\frac{b}{a} = \frac{c}{b}$$

$$\log_x \frac{b}{a} = \log_x \frac{c}{b}$$

$$\log_x b - \log_x a = \log_x c - \log_x b$$

$$\log_xa, \log_xb, \log_xc$$ are in A.P.

$$\log_a x, \log_b x, \log_c x$$ are in H.P.

50. Let $$y = (0.0504)^{10}$$

$$\log_{10}y = 10\log_{10}(.0504) = 10\log_{10}(504*10^{-4})$$

$$= -10\log_{10}[-4 + \log(2^3.3^2.7)]$$

$$= -12.98$$

Thus, chraracteristics is $$-13$$

Therefore, number of zeroes after decimal and first significant digit $$= 12$$