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# Logarithm

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Solution of Logarithm,
Book - Business Mathematics and Statistics (S.N.Dey)

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### Logarithm

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3. 3. 1. Express the following relations in the logarithmic forms: Ans. HOME Solution: 3 81 log 81 4 3 4   Logarithmic form : (i)
4. 4. 1. Express the following relations in the logarithmic forms: Ans. HOME Solution: 5 1 log 1 0 5 0   Logarithmic form : (ii)
5. 5. 1. Express the following relations in the logarithmic forms: Ans. HOME Solution: 1 5 32 2 1 5  32 2 log 2 32 5   Logarithmic form : (iii)
6. 6. 1. Express the following relations in the logarithmic forms: Ans. HOME Solution: 3 1 1 125 log 125 5 5 3     Logarithmic form : (iv)
7. 7. 1. Express the following relations in the logarithmic forms: Logarithmic form : Ans. HOME Solution: a c c b a b   log (v)
8. 8. 2. Express the following logarithmic forms in the exponential forms: log 64 6 Ans. HOME Solution: 6 2 2  64  exponential form : (i)
9. 9. 2. Express the following logarithmic forms in the exponential forms: exponential form : Ans. HOME Solution: 1 1 81 3 4 81 log 4 3     (ii)
10. 10. 2. Express the following logarithmic forms in the exponential forms: log 1 0 Ans. HOME Solution: 1 0   a a exponential form : (iii)
11. 11. 2. Express the following logarithmic forms in the exponential forms: 3 exponential form : Ans. HOME Solution: 1 1  125 25 2 125 log 25 3 2          (iv)
12. 12. 2. Express the following logarithmic forms in the exponential forms: P R log  exponential form : Ans. HOME Solution: R Q Q  P (v)
13. 13. 3. Find the value of x : log 128 Ans. HOME Solution: log 128   2 128 , 2  2 , 7 7 2 2    or or x x x x x  (i)
14. 14. 3. Find the value of x : log 81 4 log 81 4 Ans. HOME Solution: 81 x   4 4 4 or x ,  3 ,  3   or x x x  (ii)
15. 15. 3. Find the value of x : Ans. HOME Solution: log   3 x log 3 or x 1  125  5  , 1 5 , 3 3 5 5     or x x x  (iii)
16. 16. 3. Find the value of x : log 243 10 log 243 10 Ans. HOME Solution: 10 5 3 x   or x ,  3 or x , 3 1 2 10  , 3 1 5     or x x x  (iv)
17. 17. 3. Ans. HOME Find the value of x : Solution: log 49 7  log 49 7      7 49 1  2 2 , 7 7 2 1 , or or x 2    ,  2  2  4 or x x x x x  (v)
18. 18. 3. Find the value of x : log 7 5 2 10   x log 7 5 2 2 10    10 7 5 or x , 100  7  5 or , 100  5  7 x or x Ans. HOME Solution:     , 105 7 105 , or x 7 or , 15 x   ,  15    or x x x  (vi)
19. 19. 3. Find the value of x : log 0.0001 log 0.0001 10 log 10 4   x x (vii) x or x , 4log 10 Ans. HOME Solution:       , 4  log 10 1 10 10 10 10         or x
20. 20. 3. 5    Ans. HOME Find the value of x : Solution: log 0.25 4 log 0.25 4    1 2   5   5 10 x or x or x , 10 , 10 25 100 , 0.25 2 1 2 2 4 2 4                           or x x x  (viii)
21. 21. 4. Find the value of logarithms of : (i) to the base of 625 5 log 625 5 log 5 , 4log 5 Ans. HOME Solution: , 4  log 5 1 5 5 4 5     or or
22. 22. 4. Find the value of logarithms of : 343 7 log 343 7 log 7 , log 7 , log 7 6 , 6log 7 log 7 1 Ans. HOME Solution:         , 6 7 7 7 2 3 7 3 7 or or or or to the base of      (ii)
23. 23. 4. Find the value of logarithms of : to the base of   Ans. HOME Solution: 0.1 9 3   log 0.1 1 let x let 9 3   1   2 2 4 5 1 1 2  , 3 3 3 1  2 2 2  , 3 3 5       , 3 3 2 5 5 ,   or or or or or x , 2 2 2 9 , 9 3 9 log 1 9 0.1 9 log 2 2 . 9 3 . 9 3 .                                 or x x x x x   (iii)
24. 24. 4. Find the value of logarithms of : to the base of 1728 2 3 log 1728 2 3   log 2 3 6 3 let log 2  3  x 6 3 x , 2 3 2 3     x , 2 3 2 3 x , 2 3  2  3 Ans. HOME Solution:             6 6   ,    6 2 3 2 3  , 6 3 2 6 2 3 6 3 2 3      or or or or or x x  (iv)
25. 25. 4. Find the value of logarithms of : to the base of 2401 7 log 2401 7 3 log 7 let log 7 x x 3 4 , 7 7 1  3 4      , 7 7 Ans. HOME Solution:   4 1 , or or or x 3 , 12 4 7 4 7 3 3 3         or x x  (v)
26. 26. 4. Find the value of logarithms of : to the base of 2 4 Ans. HOME Solution: log 2 let x 4 8 4   , 4  2   4 8 8   2 8 , 2  2 , 2  2  or or or or x , 2   8 8 2 , log 2 2 8 -8      or x x x x  (vi)
27. 27. 4. Find the value of logarithms of : Ans. HOME Solution: to the base of 81 9 log 81 9 let log 81 x 9  3  x , 9 81 1  3 2      , 9 9 1 3 2 , 9 9 2 1 , or or or or x 3   , 2 3 6 3 3 3         or x x x  (vii)
28. 28. 4. Find the value of logarithms of : to the base of 5 0.008 log 5 0.008 let x x , 0.008 5 1    1 1 1 Ans. HOME Solution:     8 1 1         , 5 5 , 5 5 1 1 1 1 6 or or or or or or or x or x , 6 1   2 3 , - 2 , -3 5 5 , 5 125 , 5 1000 , log 5 2 -3 2 -3 2 3 2 2 3 0.008                       or x x x x x x  (viii)
29. 29. 5. Find the base when: is the logarithamof 3 343 let base x   log 343 3 3 or x ,  343 3 3 Ans. HOME Solution: or x ,  7 ,  7  or x x (i)
30. 30. 5. Find the base when: is the logarithamof 4 144 let base  x   log 144 4 4 or x ,  144 4 4 2 or x , 2 3     or x , 2 3 or x , 2 3 Ans. HOME Solution:         or x ,  2 3 , 2 3 4 4 4 4 4 2 2 4 4     or x x (ii)
31. 31. 5. Find the base when: is the logarithamof let base  x    1 1 Ans. HOME Solution:   1 3 1 1     1 or x or , x or x ,  3 , 27 3 , 3 3 log 1 3 3  3 3 3          or x x (iii)
32. 32. 5. Find the base when: let base  x    Ans. HOME Solution:   1 or , x a or x a a or x a a is the logarithamof x      , 1 , 1 1 log 1 1 1 1 (iv)
33. 33. 6. Find the simplest value of : log 5 log 27 log 27 10 log 3 3log 3 3 25 log 5 10 log 5 10 log 5   Ans. HOME Solution: 3 2 or, 10 2log 5 10 log 3 or, log 5 log 3 or, log 25 log 3 or, 10 10 2 10 3 10 10 10 10  (i) 
34. 34. 6. Find the simplest value of : log 27 if log 3 8 2 log 27 10 log 8 10 log 3 log 2 3log 3 10 3log 2 or, or, or, or, log 3 Ans. HOME Solution: a a   a  or, log 3 2 2 10 3 10 3 10  (ii)
35. 35. 6. log log log 1 2 1 x x log Ans. HOME Find the simplest value of : (iii) x x a log if log Solution: 2 2 2  1 log x log x  x 2 1 2 3 2 2 1 1 2 1 log log log log 2 or, log 2 or, log 2 or, log 2 2 or, log 2 2 or, x x   1 1 1 3 or, 3 or, log 3 or, log 2 3 or, 3log 2 or, 3log 2 or, log 2 or, 2 2 3 a a x x x    
36. 36. 7. Prove that log(1 23)  log1 log2  log3 proved HOME Solution: L H S . . . log(1  2  3) log(6) log(1  2  3) log1  log 2  log 3 or, or, or,
37. 37. 8. Express M in term of N M N log 3log 1 2 1   M N   M N or, log  log  1   or, log log 3 log 1 solved HOME Solution: given :   3 3 1 1 1 2 3 M N or,   3 2 2 3 3   3 2 6 1 2 3 2 3 3 3 3 2 3 3 3 3 9 or, 3 or, 3 or, log 3log 1 2 1     N N M N M N M M N a a              (i)
38. 38. 8. Express M in term of N (ii)   N M log 3 2log 10 10 N   M N M or, log  log  3 N M or, log (  )  3 N M or, log (  )  3  log 10 or, log ( ) log 10 2 3 or,   10 or,   1000 1000 solved HOME Solution: given : N M N M N M N M N M 1000 or, l og 3 log 2 2 3 10 2 10 10 2 10 2 10 2 10 10 2 10 10     
39. 39. 9. Prove that : log log x a  x let y a x y a   log log . ..(i) log x x a a a log log a a x a     log 1 [ log 1] a a y x log log a a y x x a x proved HOME Solution: or, log or, a f rom (i) a a a      (i)
40. 40. 9. Prove that : a 2log 2 2log x  a let y x a y x   log log . ..(i) 2log a a x x x 2log log x x a x     2log 1 [ log 1] x x y a log 2log x x y a or, log log x x 2 y a a 2log 2 x a proved HOME Solution: or, 2 or, x f rom (i) x x x       (ii)
41. 41. 9. y x log log log y x y (iii)  z x   log log . ..(i) log y a a a y x log log a a log log   z y log log a a a a log x a z y y x log log proved HOME Prove that : Solution: or, log x or, a a f rom (i) a a a a x y x y let z x    
42. 42. 9. m n n m (iv)    log log log log a b a b n b m b L H S m  n a b m a n a   n m R H S proved HOME Prove that : Solution: log log log log . . . log log log log log log or, or, or, log  log  . . . a b
43. 43. 9. log 3  log 2  1 1 2 3 . . . log 3 log 2 proved HOME Prove that : Solution: 2 3 or, 1 . . . ] 1 log [ log log 3 or, log 3 2 2 R H S a b L H S b a      (v)
44. 44. 9. b c a log  log  log  1 a b c . . . b c a log  log  log log c proved HOME Prove that : Solution: log b or, 1 . . . ] log log [ log log log log log or, R H S b a b a c b a L H S a a b c      (vi)
45. 45. 9. Prove that :     2 2 log log log( ) log . . . x y log log x   x y x y a b a b a b or, log  log log  log [   (  )(  )] x   proved HOME Solution:        or, log( )log . . . 2 2 2 2 R H S y xy L H S y x y xy                  (vii)
46. 46. 10. Find the value of : (i)  log 729 9 (27) 4 3 log 729 9 (27) 4 4  or, log 3 3 (3 ) 4 6 3  2  4 3 or, log 3 3 3 or, log 3 3 4 6 2 3 or, log 3 or, log 3 or, log 3 4 6 4 solved HOME Solution: given :         or, 1 3 4 3 4 3 6 3 4 3 3 6 2 1 3 3 3 1 3 4 3 3 1 3           
47. 47. 10. Find the value of : log log log 81 3 2 3 log log log 81 3 2 3 or, log log log 3 [  3  81] or, log log 8log 3 [ log 1] 3 2 3 or, log log 2 or, log 3 solved HOME Solution: given :       or, 1 3 3 3 2 8 8 3 2 3  a a  (ii)
48. 48. 10. Find the value of : log 16  log 9 2 3 log 16 log 9 2 3 or, log 2  log 3 [  2  16 3  9] or, 8log 2 4log 3 [ log 1] solved HOME Solution: given :             8 4 or, or, 2 2 3 4 8 4 3 8 2    a and a  (iii)
49. 49. 10. Find the value of : b c d a (iv)    log log log log a b c d b c d a log  log  log  log a b c d log b solved HOME Solution: given : or, 1 log log log d log log log log or, a d c c b a   
50. 50. 10. Find the value of : (v)   log 27 log 8 log 1000 log1.2 log 27  log 8  log 1000 3 3 3 log 3  log 2  log 10 12 3     log 3 log 2 log10 3 3 log10 2    log 3 3log 2 2     log 3  2log 2  log10   log 3  2log 2  log10 solved HOME Solution: given :     3 3     3 2 or, log 3 2log 2 log10 2 or, log 3 2log 2 log10 2 or, log 3 2 log10 3 or, log12 log10 or, 10 log or, log1.2 2 2 2 3  
51. 51. 10. Find the value of : (vi)     log 4 log 5 log 6 log 7 log 3 3 4 5 6 7 log 4  log 5  log 6  log 7  log 3 log 4 solved HOME Solution: given : or, 1 log 3 log 7 log 7 log 6 log 6 log 5 log 5 log 4 log 3 or, 3 4 5 6 7    
52. 52. 10. Find the value of : log 6 6 6... x let  6 6 6...  . ..(i) 2 x or, 6 6 6 6...   x x or,  6 from (i) or, 6    log 6 6 6... log 6 6 solved HOME Solution: given : or, log 6 1 6 2 6    x x (vii)
53. 53. 11. 25     25    16 12 7  5 25 5 16 16  16 2 2 25 16 . . .  24     64           or, log 2 5 3 81 81   3 3  81 81  28 7        proved HOME Prove that: Solution:     or, log 2 5 1 or, log10 or, 1 . . . 2 5 2 3 3 5 or, log 2 2 5 2 3 3 5 or, log 2 80 24 15 or, log 2 80 log 24 log 15 or, log 2 log 80 7 log 24 12log 15 log 2 16log 1 80 7 log 24 12log 15 log 2 16log 1 64 36 28 24 16 7 28 16 12 28 36 12 16 16 7 4 4 12 3 2 16 4 16 12 7 R H S L H S                                                                           (i)
54. 54. 11.      7 2 3 25  25        7 2 5 5 4  6 2 2 3 10  10    5 2  7 7 5 2 7 7 5 2 25 25 10 . . . 10    12 12  7 4 3 7 6 12 2 12 14        or, log 5 2 3 81 81 81 3 3   proved HOME Prove that: Solution:    0 1 0    or, log 5 2 3 or, log 1  2  1 or, log2 . . . 81  2 5 5 3 or, log 2 5 2 3 3 or, log 3  2 5 2 3 3 or, log 80 24 9 or, log 80 log 24 log 9 or, log 80 3log 24 2log 9 7 log log 2 80 3log 24 2log 9 7 log 1 12 3 4 14 12 3 6 2 14 3 4 4 2 3 2 7 2 3 R H S L H S                                                                              (ii)
55. 55. 11.      7 5 3 25  25     10 5 16  16 2 2 5 81 81 81 25 25 16 . . . 16     12    28      28 15 12 10 7 3 12 7 5       or, log 2 5 3 proved HOME Prove that: Solution: 3 81       or, log 2  5  3 or, log 2  1  1 or, log2 . . . 3 2 5 2 3 3 5 or, log 2 5 2 3 3 5 or, log 80 24 15 or, log 80 log 24 log 15 or, log 80 3log 24 5log 15 7log log 2 80 3log 24 5log 15 7log 1 0 0 12 3 15 5 7 7 3 4 4 5 3 2 7 4 7 5 3 R H S L H S                                                                        (iii)
56. 56. 11. 32    32  5 2 2 2         5 3 75 5 5    3 5 3 5 3 5 5 75 . . . 75    5 5 2 4  2 5 2   1 4 5 2 2 5 4     or, log 3 5 2 2 32    proved HOME Prove that: Solution:       or, log 3  5  2 or, log 1  1  2 or, log2 . . . 3 5 2 or, log 3 3 2 or, log 3 3 2 or, log 243 log 9 log 16 or, log 243 log 9 2log 16 log log 2 243 log 9 2log 16 log 0 0 1 5 2 4 2 5 4 4 2 5 4 2 R H S L H S                                                    (iv)
57. 57. 11. (v) x y z y z z x x y log  log log  log log  log    y z z x x y log  log log  log log  log let k  x  y  z k x y z or, log log 1 y z z x x y log  log log  log log  log    y z z x x y log  log log  log log  log k x y z or, log  log  log  log k y z x z x y x y z or, log  log  log log  log  log log  log  log log k y x z x z y x y x z y z or, log  log log  log log  log log  log log  log log  log log proved HOME Prove that: Solution:         k or, log  0 k or, log log1 k or, 1 y z z x x y log log log log log log or,    1      x y z
58. 58. 11. Prove that: 1    xy yz zx 1 1 . . . 1 xyz xyz xyz xy yz zx    xy yz zx or, log ( ) xyz xyz or, log ( ) 1 log ( ) xyz or, 2log log 1 proved HOME Solution:   or, 2 log 1 log or, log log log 1 log ( ) log ( ) 2 log ( ) log ( ) log ( ) 2            xyz a a b xy yz zx L H S xyz xyz xyz xyz a b a xyz xyz xyz   (vi)
59. 59. 11. a b c    3 3 3 b c a . . . a b c log log log 3 3 3 c b    1 log log b proved HOME Prove that: Solution: log a log 1 1 27 or, 3 a 1 3 3 or, log log c 3log 3log 3log or, log log log log log log or, 1 27 log log log 3 3 3             a c b a b a a c b L H S b b c a  (vii)
60. 60. 11. . . . 2 3 a a a a log  log  log  ...  log n a a a n a or, log  2log  3log  ...  log or, 1  2  3  ...  log proved HOME Prove that: Solution:             (  1)       ( 1) 2 log 2 or, log 2 1 log log log ... log 2 3 n n a sum of natural number n n n a L H S a n n a a a a n  (viii)
61. 61. 11. x y z x y z (ix)      log log log log log log a b c b c a proved HOME Prove that: Solution: z c z a . . . x y z log  log  log a b c y   b y   c x a x b log log log log log log x y z L H S log log log or, log log log b c a or, log log log or,  
62. 62. 11. Prove that: x y z log  log  log   1 1 1 1 y z x . . . x y z log  log  log 1 1 1 log y   log y log 1 x x 1 log log log proved HOME Solution: 1 or, 1 1 1 1 or, z log z log 1log log 1log 1log or, log log log or, 1 log 1 log 1 log or, 1 1 1                 x z z y y x x z y x z y L H S y z x (x)
63. 63. 11. Prove that: x y z    2 2 2 x y z . . . x y z log log log 2 2 2 x y z    1 1 1 1   proved HOME Solution:    1 1 8 or, log 1 2 2 2 or, 1 1 1 2log 1 1 2log 2log or, 1 log log log log log or, 8 log log log 2 2 2           a x y z b a x y z L H S a x y z a b x y z   (xi)
64. 64. 12 (a) log 25 8 log 25 log 8 log 5 2 log 2 3 2log 5 3log 2 10 2 2log  3log 2  2 log10 log 2 3log 2 2 1  0.3010 2 0.6990 solved HOME log 2 0.3010 , log 25. 8 If  find the valueof       0.9030 1.3980 0.9030 1.548 log 2 0.3010 given 3 0.3010 Solution:                
65. 65. 12 (b) log 3 , log 5 , log 8. 30 30 30 If  a and  b find the valueof  30  log 8 log 2 3log 2  30 30               3 log 30 log 15 30 30    3 log 30 log 3 5 30 30    31 log 3 log 5 30 30    31 log 3 log 5    a  b    3 1 15 3log 30 30 30 3 30 solved HOME  log 3 and log 5 given 30 30   a  b Solution:
66. 66. 13. (i) 1 1   If a 2  b  2  7 ab , show that, log a  b  log a  log b a  b  ab 2 2 a b ab ab ab or,   2  7  2 2     a b ab or,   9       ab ab  a b 1 1 1         a b ab 1  a b  a b a b ab a b ab a b log log 2 1 1 ( ) 3 3 or, or, log log 2 ( ) 3 or, log log 3 or, log 3 or, 9 or, 7 2 2 2 2 2 2                    2 proved HOME 3    Solution: given : [ Adding 2ab on both sides ] [ a2+b2+2ab=(a+b)2 ] [ taking log on both sides]
67. 67. 13. (ii) 1 y x log log  , 23 2  log             1   2 2 25  2 or, 2 2 x y xy xy or,   2  25 2 2 x y xy xy or,   25  2 2 2 x y xy or,   23 y x or,   23 y x or, 23 log 5 or, log log log 5 or, 2log log log 2 5 log 2 2 2             x y xy xy xy x y xy xy x y x y x y x y x y proved HOME 5 x y x y showthat x y If Solution: given :
68. 68. 13. (iii)   If a b a b showthat x x x x x , log log 3 5 5 3   x x x x 3  5 5  3    x x 5 3 5 3   x x x x 5  3 5   (3  )  x x x x a a b a b a b b a 5 3 5 3     b a x x 2 2 2 b a or, or, or, or, or, or, or, 2  2 2 2 2         or, log log b a b x a b a a x b b a a a a a a b x x x x x x   or, log log               a b a    proved HOME      Solution: given :
69. 69. 13. (iv) 4 4 2 2 a  b  14 a b 4 4 2 2 2 2 2 2 a b a b a b a b or,   2  14  2       2 2 2 2 2 a b a b or,   16 2 2 2 2 a b ab or,   4 2 2 a b ab or,   4  a 2 b 2   ab        or, log   log 4 e e 2 2 a b a b or, log   log 4  log  log e e e e 2 2 2 a b a b or, log   log 2  log  log e e e e 2 2 a b a b or, log   2log 2  log  log e e e e or, log  a 2  b 2   log a  log b  2log 2 e e e e proved HOME 14 , log   log log 2log 2 4 4 2 2 2 2 e e e e If a  b  a b showthat a  b  a  b  Solution: given : [ Adding 2a2b2 on both sides ] [ a4+b4+2a2b2 =(a2+b2)2 ] [ taking log on both sides]
70. 70. 14. (a) y y x x log log log       x k y z    z z log . ... .(i) y k z x log   . ... .(ii) z k x y log . ... .(iii)         from (i), (ii) and (iii) : x y z k y z k z x k x y l og log log           or, log x  y  z  ky  kz  kz  kx  kx  ky xyz or, log( )  0 or, log( ) log1 or,  1        xyz xyz k x y z x y z let HOME , 1 log log log       prove that xyz x y z x y z If Solution: proved [ log1=0]
71. 71. 14. (b) (i) x y x log log log y z z             x k b c a x ak b c log log . ... .(i)       y k c a b y bk c a log    log   . ... .(ii) z k a b c z ck a b log    log   . ... .(iii)       from (i), (ii) and (iii) : a log x  b log y  c log z  ak b  c  bk c  a  ck a  b a b c x  y  z  akb  akc  bkc  bka  cka  ckb or, log log log a b c x y z or, log(   )  0 a b c x y z or, log( ) log1 or,  1        a b c x y z k a b c a b c let HOME , 1 log log log       a b c provethat x y z a b c a b c If Solution: proved [ log1=0]
72. 72. 14. (b) (ii) y x y x log log log  z z          2 2                      x k b c b c x b c k b c k b c           log log . ... .(i) 2 2 y k c a c a y c a k c a k c a log     log      . ... .(ii) 2 2 z k a b a b z a b k a b k a b log     log      . ... .(iii) from (i), (ii) and (iii) :  b  c  log x   c  a  log y   a  b  log z  k  b  c   k  c  a   k  a  b   b  c   c  a   a  b                  x y z kb kc kc ka ka kb or, log log log         b  c c  a a  b x y z or, log    0 b  c c  a a  b x y z or, log log1       or, 1 2 2 2 2 2 2 2 2 2 2 2 2        b  c c  a a  b x y z k a b c a b c let HOME , 1 log log log       bc ca ab provethat x y z a b c a b c If Solution: [ log1 = 0] proved
73. 73. 14. (b) (iii) y x y x log log log z z             x k ry qz p x pk ry qz       log log . ... .(i) y k pz rx q y qk pz rx log    log   . ... .(ii) z k qx py r z rk qx py log    log   . ... .(iii)       from (i), (ii) and (iii) : p log x  q log y  r log z  pk ry  qz  qk pz  rx  rk qx  py p q r x y z kpry kpqz kqpz kqrx krqx krpy or, log log log          x p y q z r    or, log    0 p q r x y z or, log log1 or,  1        p q r x y z k qx py pz rx ry qz let HOME , 1 log log log       p q r prove that x y z qx py pz rx ry qz If Solution: [ log1 = 0] proved
74. 74. 15. log 45 2   log(3 5) 2   log 3 log 5   2log 3 log 5 10      2log 3 log10 log 2     2 0.4771 1 0.3010   0.9542 0.6990 1.6532 2 2log 3 log        HOME If log 2  0.3010, log3  0.4771, log 7  0.8451, f ind the value of : Solution: (i) solved [ log10 = 1]
75. 75. 15. log108 2 3   log(2 3 ) 2 3   log 2 log 3   2log 2 3log 3     2 0.3010 3 0.4771   0.6020 1.4313 2.0333  HOME If log 2  0.3010, log3  0.4771, log 7  0.8451, f ind the value of : Solution: (ii) solved
76. 76. 15. log 84 2    log(2 3 7) 2    log 2 log 3 log 7    2log 2 log 3 log 7     2 0.3010 0.4771 0.8451     0.6020 0.4771 0.8451 1.9242  HOME If log 2  0.3010, log3  0.4771, log 7  0.8451, f ind the value of : Solution: (iii) solved
77. 77. 15. log 294    log(2 3 7 )    log 2 log 3 log 7    log 2 log 3 2log 7     0.3010 0.4771 2 0.8451     0.3010 0.4771 1.6902 2.4683 2 2  HOME If log 2  0.3010, log3  0.4771, log 7  0.8451, f ind the value of : Solution: (iv) solved
78. 78. 15. log 21.6 216 10 log     log 216 log10  3 3     log 2 3 log10 3 3    log 2 log 3 log10    3log 2 3log 3 log10      3 0.3010 3 0.4771 1     0.9030 1.4313 1 1.3343       HOME If log 2  0.3010, log3  0.4771, log 7  0.8451, f ind the value of : Solution: (v) solved [ log10 = 1]
79. 79. 16 (i) given If three positive real numbers a b and c are inG P , . ., solved showthat a b and c are in A P a, b and c are in G.P. c b   b c   log log            b a c b log  log  log  log a b c A P. a b b a hence, log , log and log are in . HOME log , log log . . Solution:
80. 80. 16 (ii) log log x k log   log log log    log . ... .(i) y k log  2 . ... .(ii) z k log  3 . ... .(iii) from (i), (ii) and (iii) : y x k k k log  log  2   . ... .(iv) z y k k k log log 3 2 . ... .(v) from (iv) and (v) : y x z y     log log log log y z z    Hence, x, y and z are in G.P. y y x y x k x y z let                 log log 3 2 1 HOME , , , . . 3 2 1 provethat x y z are inG P x y z If   Solution:
81. 81. 17. The f irst and the last terms of aG P are a and k respectively If the number of termbe n provethat k a log log given First term = a Last term = k No. of term = n Common ration = r . . .  Last terminG P ar 1 k ar log  log k a r log  log  log k a n r log  log   1 log k a n r log log 1 log proved HOME . log 1 , , where r is the commonratio r n    Solution:          n k a log log k a log log r n r k ar n n n n             1 log 1 log . . 1 1 1
82. 82. 18. If the p q and r terms of aG P are a b and c respectively showthat th th th given , . . , , q  r a  r  p b  p  q c  pth term = a qth term = b rth term = c let the first term = b and common ratio = R pth term = a = b Rp-1 qth term = b = b Rq-1 rth term = c = b Rr-1 Press ( ) log ( ) log ( ) log 0 Solution:
83. 83. since a, b and c are in G.P. - - - (ii) - - - (iii) - - - (i) b q p      R 1 1 1 R  1 q 1 1 1 1 1 1 R b c b b   and b c b R R R c b R a R R R a b a r r q r q r q q q p q p p                          b b  Press
84. 84. from (i), (ii) and (iii) : 1 1 b q p c r q    c       q  p q p b r q c b  b     r  q q  p r  q q  p b  b  a  c r  q  q  p r  q q  p b  a  c r  p r  q q  p b  a  c r p r q q p r q r q q p b a c a b a b a                   log log log Press
85. 85. r p b r q a q p c (  ) log  (  ) log  (  ) log r p b q r a p q c (  ) log   (  ) log  (  ) log q r a p q c r p b (  ) log  (  ) log  (  ) log  0 q r a r p b p q c (  ) log  (  ) log  (  ) log  0 solved HOME
86. 86. 19. (i) If x  bc y  ca and z  ab showthat a b c 1 x y 1 z 1 x y z 1 1 bc ca ab a b c bc a ca b ab c a a b b c c 1 1 1 1 1 . .    abc abc abc a b c a b c    log log log log 1 1 1 1 log ( ) log ( ) log ( ) 1 log ( ) log log ( ) log log ( ) log log ( ) 1 log ( ) 1 log ( ) 1 1 1 1 1 1 1                    abc abc abc abc L H S abc proved HOME 1 1 ( ) log ( ), log ( ) log ,       i Solution:
87. 87. 19. (ii) given If x  log ( bc ), y  log ( ca ) and z  log ab , showthat a b c ( ) 2 x bc log ( ) a  y ca log ( ) b  z ab log ( ) c  hence 1 1 1 bc ca ab a b c x y z x y z y x    1 1 x y     1 1 ( 1) ( 1) x y z z xy x y (   2)(  1)  (    1) xz  x  yz  y  z   xyz  xz  yz  z x  y  2 z  2  xyz  z x  y  2 z  z  2  xyz x y z xyz z z xy x y z z x y                                2 2 2 2 ( 1) ( 1)( 1) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 log 1 log 1 log 1 proved HOME ii x  y  z   xyz
88. 88. 20. given a a a . . . 1     log log 2 log 3 1 a a a 2 3 4 a     a L H S log log log 2   a a log log 4 a log 4 a a log( 4 ) log 4 a log(2 ) a a a 2log 2 log 4 log 2 a       a a log 3 log 2 log 3 log 4 3    2 log 2 log 3 proved x a a 2  y log 2 a a 3  z log 3 a a log 4  yz a a a a a a a a a a a a a xyz a a a 2 log 4 2 log 3 2 log 4 1 log 4 1 log 4 log 3 log 2 3 4 3 2       HOME If x a y a and z a showthat xyz yz a a a log , log 2 log 3 , 1 2 2 3 4     
89. 89. 21. log  , log  , prove log  p q p  given p q 1 p q . . . 1 x log log log x x 1 ab proved x  a x b p q  log log L H S p 1 1 1 1 a b b a ab ab b a x x x p q q            1 log log 1 log HOME b a If x a and x b that x q
90. 90.   If log x y a, and log , log log 2 3    22. Solution: given: x log( x 2 y 3 ) a - - - (i) b x   log - - - (ii) y         solved (i) from 2 3 x y a log log   x y a 2log  3log  - - - (iii) x y b log log   log x  log y  b - - - (iv) from (iii) and (iv)   y b y a 2 log 3log    y b y a 2log 2 3log    y a b 5log 2 2 5 log (ii)   a b y from    and from a 2 b b 2 5 3 a b 2 3 5 log 5  log  5 5 5 log (iv) a b y a b x a b b x            HOME   b prove that x and y in terms of a and b y     
91. 91. 23. e y 2  2  solved y y e  e 1 If x  y  y e y y e e   y y y y e e   y y e e e e y  y y  y  e  e  e  e e e y y y e y y y y e e   y y e e y y y y  likewise   e  e  e  e y y 1 - - - (ii) e 2 1 1 - - - (i) 2 1 1 e e y y y e e x e e x x                         y y y y     1 1 e e x x from and x y y e e e e x x y e y x x x e x e e e e e x e e y e e y y y y y y   1 1                    1 1 1 1 1 log 2 log 2 2 log 2 1 log log 1 log 2 2 1 (i) (ii) 2 2 HOME x x show that y e e      1 log 2 log , Solution: given:
92. 92. 24. log 3 10 Showthat the valueof lies between and Solution: HOME 2 . 5 1 2 9 10 log 9 log 10 10 10 - - - (i) log 3  1 2log 3 1 1 2 log 3 10 10 2 10      and 243 100 log 243 log 100 10 10 log 3  log 10 5log 3  2log 10 10 10 - - - (ii) 5log 3  2 2 5 10 log 3 10 2 10 5 10     from i and ii 2 5 log 3 1 2 ( ) ( ) 10   proved
93. 93. 25. Solution: HOME log 10 log (32 ) 5, . 6 If b and b find the value of a a a   a  log 10 (given) 10 a b log (32 ) 5 (given) 5 6 a b (6 )  32 - - - (ii) from (i) and (ii) 5 10  b a (6 ) 32  6 2 5 6 2   3 1 1 2 10 5 5 10 5 a 6 32 6 - - - (i) 5 5 5 5 5 5 a             a a a a b a b and b a found
94. 94. 26. Solution: HOME found
95. 95. 27. (i) Solve Solution: HOME 4 solved x x x 2 10 log  log  10 10 log 2 x  x   10 1 x x x x x x  x x x x x 2 2 2 10 1 2 10 10   1 2 1 10 2 10 10 10 2 10 10 10 log log log log log log log log log log log                                    2 2 2 1 x x log  log  4 10 10 x (log ) 4 x log  4   2 1  or , 10 x 2 hence x or 10 10 100 100 log log log log log 2 2 2 10 2 10 10 10 10 10 10  x or x x x x x          
96. 96. 27. (ii) Solve Solution: log log log 1 2 2 2 x   x log log log 1 2 2 2   log log log log 2 2 2 2 2 log log 2 2 2  x log log  2  log 2 2 2 2 x log log  log 2 log log log 4 2 2 2 HOME solved log 4  2  2 4 , 16 2 2 2 2     or x x x x again x x
97. 97. 27. (iii) Solve Solution: HOME log log log 11 8 4 2 x  x  x  log  log  log  11 8 4 2 1     x x x 1    x x x    1 1 1     2 3 6 6 1 1  solved 1 1 1 1 x 1 x 1 x 1    x log 6 , 64 2 6 log 2 11 11 log 2 11 6 log 2 1 11 2 3 log 2 11 1 log 2 2log 2 3log 2 11 log 2 log 2 log 2 11 log 2 log 4 log 8 6 2 3 2             or x x x x x x x x x 
98. 98. 28. If a b and c are three consecutive positive integers showthat log(1 ) 2log Solution: HOME , ,   a b and c are three consecutive positiveintegers proved ac b b ac   1  , 2 2 b ac log  log(1  ) b ac 2log  log(1  ) 