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Mathematics

Mathematics. Session. Logarithms. Session Objectives. Session Objectives. Definition Laws of logarithms System of logarithms Characteristic and mantissa How to find log using log tables How to find antilog Applications. Logarithms Definition. Base: Any postive real number

julinka-horvath
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Mathematics

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  1. Mathematics

  2. Session Logarithms

  3. Session Objectives

  4. Session Objectives • Definition • Laws of logarithms • System of logarithms • Characteristic and mantissa • How to find log using log tables • How to find antilog • Applications

  5. Logarithms Definition Base: Any postive real number other than one 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 log27 is (a) Integer (b) Rational (c) Irrational (d) Prime Solution: Log27 is an Irrational number As there is no rational number, 2 to the power of which gives 7 Why?

  7. Fundamental laws of logarithms

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

  9. Illustrative Example Solution:

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

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

  12. Illustrative Example Solution:

  13. Illustrative Example Solution:

  14. Illustrative Example If log32, log3(2x-5) and log3(2x-7/2) are in arithmetic progression, then find the value of x Solution: 2log3(2x-5) = log32 + log3(2x-7/2) Why log3(2x-5)2 = log32.(2x-7/2) (2x-5)2 = 2.(2x-7/2) 22x -12.2x + 32 = 0, put 2x = y, we get y2- 12y + 32 = 0  (y-4)(y-8) = 0  y = 4 or 8  2x=4 or 8  x = 2 or 3

  15. If a2+4b2 = 12ab, then prove that log(a+2b) is equal to Illustrative Example Solution: a2+4b2 = 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 Log10x, also known as Brigg’s system Note: if base is not given base is taken as 10 Natural logarithm: Base = e Logex, also denoted as lnx Where e is an irrational number given by

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

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

  19. How to find log(n) using log tables 1) Step1: Standard form of decimal n = m x 10p , 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 Example n = m x 10p, 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) Never neglect 0’s at end or front Summarize: 1) Std. Form n = m x 10p 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)

  23. Illustrative Example Find log(1234.56) 1) Std. Form n = 1.23456 x 103 2) Significant digits of m: 1234 3) Number at (12,3) = 0899 Note this 4) Number at (12,4) = 14 5) log(n) = 3 + .(0899+14) = 3 + 0.0913 = 3.0913

  24. To avoid the calculations Illustrative Example Find log(0.000123) 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

  25. Illustrative Example Find log(100) 1) Std. Form n = 1 x 102 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. To avoid the calculations Illustrative Example Find log(0.10023) 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

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

  28. Eg. n = -1.2718 How to find Antilog(n) 2) Step2: Select row ‘ab’ Select the row ‘ab’ from the antilog table 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. If i.e. n < 0 Antilog(n) = .(x+y) x 10-(m-1) How to find Antilog(n) 5) Step5: Antilog(n) If n = m.abcd i.e. n  0 Antilog(n) = .(x+y) x 10m+1 y = 2 x = 5346 Antilog(n) = .(5346 + 2) x 10-(2-1) = .5348 x 10-1 = 0.05348

  31. Illustrative Example Find Antilog(3.0913) Solution: 1) Std. Form n = 3.0913 = m.abcd 2) Row 09 3) Number at (09,1) = 1233 4) Number at (09,3) = 1 • Antilog(3.0913) • = .(1233+1) x 103+1 • = 0.1234 x 104 = 1234

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

  33. Illustrative Example Find Antilog (2) Solution: 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) = .(1000+0) x 102+1 = 0.1000 x 103 = 100

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

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

  36. Find Illustrative Example Solution:

  37. Solution Cont. = 0.2708 x = antilog (0.2708) = 0.1865 × 101 = 1.865

  38. Illustrative Example Find the compound interest on Rs. 20,000 for 6 years at 10% per annum compounded annually. Solution: = 20000 (1.1)6 logA = log [20000 (1.1)6] = log 20000 + log (1.1)6 = log (2 × 104) + 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 × 105 = 35430 Compound interest = 35430 – 20000 = 15,430

  40. Illustrative Example 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. Solution: = 80000 (1.075)2 log p2 = 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 p2= 4.9659 p2 = antilog (4.9659) = 0.9245 × 105 = 92450

  42. Illustrative Example 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. Solution: = 6250 (0.98)3 log v2 = log 6250 + 3 log 0.98 = log (6.250 × 103) + 3 log (9.8 × 10–1) = log 6.250 + 3 + 3 log (9.8) – 3 = 0.7959 + 3 × (0.9912)

  43. Solution Cont. log v2= 0.7959 + 3 × (0.9912) = 3.7695 v2 = antilog (3.7695) = 0.5882 × 104 = Rs. 5882

  44. Class Exercise

  45. Find Class Exercise - 1 Solution :

  46. If a2 + b2 = 7ab, prove that Class Exercise - 2 Solution : a2 + b2 = 7ab a2 + b2 + 2ab = 9ab (a + b)2 = 9ab taking log both sides we get

  47. Find x, y if Class Exercise - 3 Solution : logx = 2 log5 = log52 = log25 Similarly x = 25 y = 8

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

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

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

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