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1. Ratio and Proportion, Indices and
Logarithm– Chapter 1
Paper 4: Quantitative Aptitude-Statistics
Ms. Ritu Gupta B.A. (Hons.) Maths and MA (Maths)

2. Introduction to Logarithm
• Fundamental Knowledge
• Its application
2

3. Definition of Logarithm
3

4. Example
4

5. Properties of Logarithm
5

6. Things to Remember
6

7. Laws of Logarithm
7

8. Logarithm of a Product Rule
Logarithm of the product of two numbers is equal to the
sum of the logarithm of the numbers to the same base, i.e.
loga (mn) = logam + logan
8

9. 9
Logarithm of a Product Rule Contd…

10. Logarithm of a Quotient
Logarithm of a quotient of any two postive numbers to any
real base (>1) is equal to the logarithm of the numerator –
logarithm of the denominator to the same base i.e.
loga (m/n) = logam - logan
10

11. 11
Logarithm of a Quotient Contd…

12. Logarithm of a power of a number
The logarithm of a number to any rational index, to any real
base (>1) is equal to the product of the index and the
logarithm of the given number to the same base i.e.
logamn = nlogam
12

13. 13
Logarithm of a power of a number
Contd…

14. Change of Base
14

15. 15
Change of Base Contd…

16. Base Changing Result
16

17. Systems of Logarithm
17

18. Common Logarithms
Natural Logarithms
18
Systems of Logarithms

19. Natural Logarithms
•The logarithm to the base e; where e is the sum of infinite
series are called natural logarithms (e=2.7183 approx.).
•They are used in theoretical calculations
19

20. Common Logarithm
• Logarithm to the base 10 are called common logarithm.
• They are used in numerical (Practical) calculations.
• Thus when no base is mentioned in numerical
calculations, the base is always understood to be 10.
20

21. Example
Power (+) of 10
(Positive
Characteristic)
Logarithmic
Form
Power (-) of 10
(Negative
Characteristic)
Logarithmic
Form
101=10 log1010 =1 10-1 = 0.1 log100.1 =-1
102=100 log10100=2 10-2= 0.01 log100.01 =-2
103=1000 log101000=3 10-3= 0.001 log100.001 =-3
21

22. Standard form of a number n
• Any positive decimal or number say ‘n’ can be written in
the form of integral power of 10 say 10p (where p is an
integer) and a number m between 1 and 10.
• Therefore n = m x 10p
• where p is an integer (positive, negative or zero) and m is
such that 1≤m<10. This is called the standard form of n.
Example- Write the Standard Form for the following
(1) 259.8 (2) 25.98 (3) 0.2598 (4) 0.02598
22

23. Example – Continued
23

24. Characteristic and Mantissa
• The logarithm of a number consist of two parts, the whole
part or integral part is called the characteristic and
decimal part is called Mantissa.
• Mantissa is always positive and always less than 1.
• The characteristic is determined by bringing the given
number n to the standard form n=m x 10p, in which p (the
power of 10) gives the characteristic and the mantissa is
found from the logarithmic table.
24

25. Example
25

26. Rules to find Characterstic
26

27. Rule 1
• The characteristic of the logarithm of any number greater
than 1 is positive and is one less than the number of digits
to the left of the decimal point in the given number.
Example: Consider the following table
27
Number Characteristic
48 1
3578 3
8.31 0

28. Rule 2
• The characteristic of the logarithm of any number less
than 1 is negative and numerically one more than the
number of zeros to the right of the decimal point. If there
is no zero then obviously it will be -1.
Example: Consider the following table
28
Number Characteristic
.6 -1
.09 -2
.00657 -3
.000852 -4

29. Mantissa
• The Mantissa of the common logarithm of a number can
be found from a log-table.
29

30. What is Log Table
30

31. How to use the Log Table to find
Mantissa
1. Remove the decimal point from the given number.
2. Consider the first two digits.
3. In horizontal row beginning with above two digits, read
the number under column headed by 3rd digit (from the
left) of the number.
4. To the number obtained above, add the number in the
same horizontal line under the mean difference columns
headed by 4th digit (from the left) of the number.
5. Then pre-fix the decimal point to the number obtained in
4th point above.
31

32. Example
• Suppose we have to find the log 125.6
• Here characteristic is 3 – 1 = 2
• For Mantissa, which is the positive decimal part.
• First remove decimal point, number becomes 1256
• The first two digits are 12, the third is 5 and fourth is 6
32

33. Example- Continued
Mantissa = 0.(0969+ 21)
= 0.0990
log 125.6 = 2 + 0.0990
= 2.0990
33

34. Point to remember
34

35. Point to remember- Continued
35

36. Anti Logarithm
• The reverse process of finding the logarithm is called
Antilogarithm i.e. to find the number.
• If x is the logarithm of a given number n with given base
‘a’ then n is called antilogarithm or antilog of x to that
base.
• Mathematically, if logan = x
Then n = antilog x
36

37. Example
Find the number whose logarithm is 2.0239
From the Antilog Table
For mantissa .023, the number = 1054
For mean difference 9, the number = 2
Therefore for mantissa .0239, the number = 1054 + 2
= 1056
37

38. Example- Continued
Here the characteristic is 2
Therefore the number must have 3 digits in the integral
part.
Hence antilog 2.0239 = 105.6
38

39. Illustrations
39

40. Illustration 1
40

41. Illustration 2
41

42. Illustration 2 - Continued
= 28 log 2 - 7 log 3 - 7 log 5 + 10 log 5 - 15 log 2 - 5 log 3 +
12 log 3 - 12 log 2 - 3 log 5
= (28 - 15 - 12) log 2 + (- 7 - 5 + 12) log 3 + (- 7 + 10 - 3)
log 5 = log 2. = R.H.S
42

43. Illustration 3
The value of log2 [log2 {log3 (log3 273)}] is
(a) 1 (b) 2 (c) 0 (d) None of these
Solution :
Given expression
= log2 [log2 {log3 (3log3 27 )}]
= log2 [log2 {log3(31og333)} ]
= log2[log2{log3(9log33)}]
43

44. Illustration 3 – Continued
= log2 [log2 {log3 (9X1)}] (as log3 3 = 1)
= log2 [log2 {log3 32}]
= log2 [log2 (2log3 3)]
= log2 [log2 2] = log21 = 0
44

45. Illustration 4
45

46. Illustration 4 – Continued
46

47. Illustration – 5
47

48. Illustration – 5 - Continued
L.H.S. = K (y – z) (y2 + z2 + yz) + K (z – x) (z2 + x2 +xz) + K
(x – y) (x2 + y2 + xy)
= K (y3 – z3) + K (z3 – x3) + K (x3 – z3)
= K (y3 – z3 + z3 – x3 + x3 – y3) = K. 0 = 0 = R.H.S.
48

49. Illustration – 6
49

50. Illustration – 6 – Continued
50

51. Illustration – 7
51

52. Illustration – 7 – Continued
52

53. Illustration – 8
53

54. Illustration – 9
54

55. Illustration – 10
55

56. Illustration – 10 – Continued
56

57. Illustration – 11
57

58. Illustration – 11 – Continued
loga + logb + logc = ky – kz+ kz – kx + kx – ky
log(abc) = 0
log(abc) = log1
abc = 1
58

59. Illustration – 12
59

60. Illustration – 12 – Continued
60

61. Illustration – 13
61

62. Illustration – 13 – Continued
62

63. Illustration – 14
63

64. Illustration – 14 – Continued
64

65. Illustration – 15
65

66. Illustration – 15 - Continued
66

67. Illustration – 16
67

68. Illustration – 16 - Continued
68

69. Illustration – 17
69

70. Illustration – 17 – Continued
70

71. Illustration 18
logb(a) . logc(b) . loga(c) is equal to
(a) 0 (b) 1 (c) -1 (d) None of these
Solution:
logb(a) . logc(b) . loga(c)
= logca . logac
= logaa
=1
71

72. Illustration 19
alogb – logc . blogc – loga . cloga – logb has a value of
(a) 1 (b) 0 (c) -1 (d) None of these
Solution:
Let x = alogb – logc . blogc – loga . cloga – logb
Taking log on both sides, we get
logx = log(alogb – logc . blogc – loga . cloga – logb)
= logalogb – logc + logblogc – loga + logcloga – logb
72
∴

73. Illustration 19 – Continued
73

74. Illustration 20
74

75. Illustration 20 - Continued
75

76. Thank You!
76