1. Mathematics

2. Session Logarithms

3. Session Objectives

4. 1.Definition 2.Laws of logarithms 3.System of logarithms 4.Characteristic and mantissa 5.How to find log using log tables 6.How to find antilog 7.Applications

5. Base:Any postive real number other than one Logarithms Definition Log of N to the base a is x Note: log of negatives and zero are not Defined in Reals

6. Illustrative Example The number log 2 7 is (a) Integer(b) Rational (c) Irrational(d) Prime Solution: Log 2 7 is an Irrational number Why ? As there is no rational number, 2 to the power of which gives 7

7. Fundamental laws of logarithms

8. Other laws of logarithms Change of base Where ‘a’ is any other base

9. Illustrative Example Solution:

10. Illustrative Example Solution : True / False ? Hence True

11. Illustrative Example Solution: If a x = b, b y = c, c z = a, then the value of xyz is a) 0b) 1c) 2d) 3

12. Illustrative Example Solution:

13. Illustrative Example Solution:

14. Illustrative Example Solution: If log 3 2, log 3 (2 x -5) and log 3 (2 x -7/2) are in arithmetic progression, then find the value of x 2log 3 (2 x -5) = log 3 2 + log 3 (2 x -7/2) log 3 (2 x -5) 2 = log 3 2.(2 x -7/2) (2 x -5) 2 = 2.(2 x -7/2) 2 2x -12.2 x + 32 = 0, put 2 x = y, we get y 2 - 12y + 32 = 0  (y-4)(y-8) = 0  y = 4 or 8  2 x =4 or 8  x = 2 or 3 Why

15. Illustrative Example Solution: If a 2 +4b 2 = 12ab, then prove that log(a+2b) is equal to a 2 +4b 2 = 12ab  (a+2b) 2 = 16ab 2log(a+2b) = log 16 + log a + log b 2log(a+2b) = 4log 2 + log a + log b log(a+2b) = ½(4log 2 + log a + log b)

16. System of logarithms Common logarithm: Base = 10 Log 10 x, also known as Brigg’s system Note: if base is not given base is taken as 10 Natural logarithm: Base = e Log e x, also denoted as lnx Where e is an irrational number given by

17. Illustrative Example Solution: True / False ? Hence False

18. Characteristic and Mantissa Standard form of decimal p is characteristic of n log(m) is mantissa of n log(n)=mantissa+characteristic

19. How to find log(n) using log tables 1) Step1: Standard form of decimal n = m x 10 p, 1  m < 10 Note to find log(n) we have to find the mantissa of n i.e. log(m) 2) Step2: Significant digits Identify 4 digits from left, starting from first non zero digit of m, inserting zeros at the end if required, let it be ‘abcd’

20. How to find log(n) using log tables nStd. form m x 10 p pm‘abcd’ 1234.561.23456x10 3 31.23451234 0.0001231.23x10 -4 -41.231230 1001x10 2 211000 0.100231.0023x10 -1 1.00231002 Example n = m x 10 p, p: characteristic, log(m): mantissa Log(n) = p + log(m)

21. How to find log(n) using log tables 3) Step3: Select row ‘ab’ Select row ‘ab’ from the logarithmic table 4) Step4: Select column ‘c’ Locate number at column ‘c’ from the row ‘ab’, let it be x 5) Step5: Select column of mean difference ‘d’ If d  0,Locate number at column ‘d’ of mean difference from the row ‘ab’, let it be y What if d = 0? Consider y = 0

22. How to find log(n) using log tables 6) Step6: Finding mantissa hence log(n) Log(m) =.(x+y) Log(n) = p + Log(m) Summarize: 1) Std. Form n = m x 10 p 2) Significant digits of m: ‘abcd’ 3) Find number at (ab,c), say x, where ab: row, c: col 4) Find number at (ab,d), say y, where d: mean diff 5) log(n) = p +.(x+y) Never neglect 0’s at end or front

23. Illustrative Example Find log(1234.56) nStd. form m x 10 p pm‘abcd’ 1234.5 6 1.23456x1 0 3 31.23451234 1) Std. Form n = 1.23456 x 10 3 2) Significant digits of m: 1234 3) Number at (12,3) = 0899 4) Number at (12,4) = 14 5) log(n) = 3 +.(0899+14) = 3 + 0.0913 = 3.0913 Note this

24. Illustrative Example Find log(0.000123) nStd. form m x 10 p pm‘abcd’ 0.0001 23 1.23x10 -4 -41.231230 1) Std. Form n = 1.23 x 10 -4 2) Significant digits of m: 1230 3) Number at (12,3) = 0899 4) As d = 0, y = 0 Note this 5) log(n) = -4 +.(0899+0) = -4 + 0.0899 = -3.9101 To avoid the calculations

25. Illustrative Example Find log(100) nStd. form m x 10 p pm‘abcd’ 1001x10 2 211000 1) Std. Form n = 1 x 10 2 2) Significant digits of m: 1000 3) Number at (10,0) = 0000 4) As d = 0, y = 0 5) log(n) = 2 +.(0000+0) = 2 + 0.0000 = 2

26. Illustrative Example Find log(0.10023) nStd. form m x 10 p pm‘abcd’ 0.1002 3 1.0023x10 -1 1.00231002 1) Std. Form n = 1.0023 x 10 -1 2) Significant digits of m: 1002 3) Number at (10,0) = 0000 4) Number at (10,2) = 9 5) log(n) = -1 +.(0000+9) = -1 + 0.0009 = -0.9991 To avoid the calculations

27. How to find Antilog(n) (1) Step1: Standard form of number If n  0, say n = m.abcd For bar notation subtract 1, add 1 we get If n < 0, convert it into bar notation say For eg. If n = -1.2718 = -1 – 0.2718 n = -1-0.2718=-2+1-0.2718 n = -2+0.7282 Now n = m.abcd or

28. How to find Antilog(n) 2) Step2: Select row ‘ab’ Select the row ‘ab’ from the antilog table Eg. n = -1.2718 Select row 72 from table 3) Step3: Select column ‘c’ of ‘ab’ Select the column ‘c’ of row ‘ab’ from the antilog table, locate the number there, let it be x Number at col 8 of row 72 is 5346, x = 5346

29. How to find Antilog(n) 4) Step4: Select col. ‘d’ of mean diff. Select the col ‘d’ of mean difference of the row ‘ab’ from the antilog table, let the number there be y, If d = 0, take y as 0 Number at col 2 of mean diff. of row 72 is 2, y = 2

30. How to find Antilog(n) 5) Step5: Antilog(n) If n = m.abcd i.e. n  0 Antilog(n) =.(x+y) x 10 m+1 If i.e. n < 0 Antilog(n) =.(x+y) x 10 -(m-1) x = 5346 y = 2 Antilog(n) =.(5346 + 2) x 10 -(2-1) =.5348 x 10 -1 = 0.05348

31. Illustrative Example Find Antilog(3.0913) 1) Std. Form n = 3.0913 = m.abcd 2) Row 09 3) Number at (09,1) = 1233 4) Number at (09,3) = 1 5)Antilog(3.0913) =.(1233+1) x 10 3+1 = 0.1234 x 10 4 = 1234 Solution:

32. Illustrative Example Find Antilog(-3.9101) 1) Std. Form n = -3.9101 2) Row 08 3) Number at (08,9) = 1227 4) Number at (08,9) = 3 5) Antilog(-3.9101) Solution: n = -3 – 0.9101 = -4 + 1 – 0.9101 n = -4 + 0.0899 =.(1277+3) x 10 -(4-1) = 0.1280 x 10 -3 = 0.0001280

33. Illustrative Example Find Antilog (2) 1) Std. Form n = 2 = 2.0000 2) Row 00 3) Number at (00,0) = 1000 4) As d = 0, y = 0 5) Antilog(2) = Antilog(2.0000) Solution: =.(1000+0) x 10 2+1 = 0.1000 x 10 3 = 100

34. Illustrative Example Find Antilog(-0.9991) 1) Std. Form n = -0.9991 2) Row 00 3) Number at (00,0) = 1000 4) Number at (00,9) = 2 5) Antilog(-0.9991) Solution: -0.9991 = -1 + 1 – 0.9991 = -1 + 0.0009 =.(1000+2) x 10 -(1-1) = 0.1002

35. Applications 1) Use in Numerical Calculations 2) Calculation of Compound Interest 3) Calculation of Population Growth 4) Calculation of Depreciation Now take log

36. Illustrative Example Find Solution:

37. Solution Cont. = 0.2708 x = antilog (0.2708)= 0.1865 × 10 1 = 1.865

38. Illustrative Example Solution: Find the compound interest on Rs. 20,000 for 6 years at 10% per annum compounded annually. = 20000 (1.1) 6 logA = log [20000 (1.1) 6 ] = log 20000 + log (1.1) 6 = log (2 × 10 4 ) + 6 log (1.1) = log2 + 4 + 6 log (1.1)= 0.301+ 4 + 6 × (0.0414) = 4.5494

39. Solution Cont. log A = 4.5494 A = antilog (4.5494) = 0.3543 × 10 5 = 35430 Compound interest = 35430 – 20000 = 15,430

40. Illustrative Example Solution: The population of the city is 80000. If the population increases annually at the rate of 7.5%, find the population of the city after 2 years. = 80000 (1.075) 2 log p 2 = log 80000 + 2 log 1.075 = log 8 + 4 + 2 log (1.075) = 0.9031 + 4 + 2 × (0.0314) = 4.9659

41. Solution Cont. log p 2 = 4.9659 p 2 = antilog (4.9659) = 0.9245 × 10 5 = 92450

42. Illustrative Example Solution: The value of a washing machine depreciates at the rate of 2% per annum. If its present value is Rs6250, what will be its value after 3 years. = 6250 (0.98) 3 log v 2 = log 6250 + 3 log 0.98 = log (6.250 × 10 3 ) + 3 log (9.8 × 10 –1 ) = log 6.250 + 3 + 3 log (9.8) – 3 = 0.7959 + 3 × (0.9912)

43. Solution Cont. log v 2 = 0.7959 + 3 × (0.9912) = 3.7695 v 2 = antilog (3.7695) = 0.5882 × 10 4 = Rs. 5882

44. Class Exercise

45. Class Exercise - 1 Find Solution :

46. Class Exercise - 2 Solution : If a 2 + b 2 = 7ab, prove that a 2 + b 2 = 7ab a 2 + b 2 + 2ab = 9ab(a + b) 2 = 9ab taking log both sides we get

47. Class Exercise - 3 Solution : Find x, y if logx = 2 log5 = log5 2 = log25 x = 25Similarly y = 8

48. Class Exercise - 4 Solution : If find y if x = 2.

49. Class Exercise - 5 Solution : Simplify (i) (ii)

50. Class Exercise - 6 Solution : Simplify and x = 2 k then k is (a) 0.25(b) 0.5 (c) 1(d) 2

51. Class Exercise – 7 (i) Solution : (i) If x, y, z > 0, such that evaluate x x y y z z. x logx + y logy + z logz x x.y y.z z = 1

52. Class Exercise – 7 (ii) (ii) If a, b, c > 0, such that then prove that a b b a = b c c b = c a a c Solution :

53. Solution Cont. Similarly Hence b loga + a logb = c logb + b logc= a logc + c loga loga b.b a = logb c c b = logc a a c

54. Class Exercise - 8 Solution : Find characteristic, mantissa and log of each of the following (i) 67.77(ii).0087 (i) 67.77 = 6.777 × 10 1 Characteristic = 1Mantissa = log (6.777) = 0.(8306+5) = 0.8311 log 67.77 = 1 + 0.8311 = 1.8311

55. Solution Cont. (ii) 0.087 = 8.7 × 10 –3 Characteristic = –3 Mantissa = log (8.7) = 0.(9395 + 0) = 0.9395

56. Class Exercise 9 Solution Find the antilogarithm of each of the following (i) 4.5851(ii) –0.7214 (i) Antilog(4.5851) =.(3846 + 1) × 10 5 = 38470 (ii) Antilog(–0.7214) = Antilog(–1 + 1 – 0.7214) =.(1897 + 3) × 10 0 = 0.19 Antilog(–1 + 0.2786)

57. Class Exercise - 10 Solution If a sum of money amounts to Rs. 100900 in 31 years at 25% per annum compound interest, find the sum. logP = log(100900) – 31log (1.25) = log (1.009 × 10 5 ) – 31log (1.25) = log (1.009) + 5 – 31 log (1.25)

58. Solution Cont. log P = 1.9998 P = Antilog (1.9998) = 0.9995 × 10 2 = 99.95 = 0.0037 + 5 – 31 (0.0969) = 5.0037 – 3.0039 = 1.9998

59. Thank you