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![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](https://image.slidesharecdn.com/secdch1ratioandproportionindiceslogarithmspart2-160828075219/75/Ratio-and-Proportion-Indices-and-Logarithm-Part-4-43-2048.jpg)
![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](https://image.slidesharecdn.com/secdch1ratioandproportionindiceslogarithmspart2-160828075219/75/Ratio-and-Proportion-Indices-and-Logarithm-Part-4-44-2048.jpg)
![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](https://crownmelresort.com/image.slidesharecdn.com/secdch1ratioandproportionindiceslogarithmspart2-160828075219/75/Ratio-and-Proportion-Indices-and-Logarithm-Part-4-43-2048.jpg)
![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](https://crownmelresort.com/image.slidesharecdn.com/secdch1ratioandproportionindiceslogarithmspart2-160828075219/75/Ratio-and-Proportion-Indices-and-Logarithm-Part-4-44-2048.jpg)
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Ratio and Proportion, Indices and Logarithm Part 4
The document covers the fundamentals of logarithms, including their definitions, properties, and various laws, such as the logarithm of a product and the quotient rule. It explains common and natural logarithms, provides examples, and details how to use logarithm tables to find characteristics and mantissas. Additionally, it discusses the concept of antilogarithms and includes several illustrations and examples to clarify the concepts introduced.
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Ratio and Proportion, Indices and Logarithm Part 4
- 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
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- 8. Logarithm of aProduct 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 aProduct Rule Contd…
- 10. Logarithm of aQuotient 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 aQuotient Contd…
- 12. Logarithm of apower 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 apower of a number Contd…
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- 15. 15 Change of BaseContd…
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- 18. Common Logarithms Natural Logarithms 18 Systemsof Logarithms
- 19. Natural Logarithms •The logarithmto 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 • Logarithmto 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 (+) of10 (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 ofa 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
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- 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
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- 26. Rules to findCharacterstic 26
- 27. Rule 1 • Thecharacteristic 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 • Thecharacteristic 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 Mantissaof the common logarithm of a number can be found from a log-table. 29
- 30. What is LogTable 30
- 31. How to usethe 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 wehave 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
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- 35. Point to remember-Continued 35
- 36. Anti Logarithm • Thereverse 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 numberwhose 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 thecharacteristic is 2 Therefore the number must have 3 digits in the integral part. Hence antilog 2.0239 = 105.6 38
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- 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 valueof 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
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- 46. Illustration 4 –Continued 46
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- 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
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- 50. Illustration – 6– Continued 50
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- 52. Illustration – 7– Continued 52
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- 56. Illustration – 10– Continued 56
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- 58. Illustration – 11– Continued loga + logb + logc = ky – kz+ kz – kx + kx – ky log(abc) = 0 log(abc) = log1 abc = 1 58
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- 60. Illustration – 12– Continued 60
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- 62. Illustration – 13– Continued 62
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- 64. Illustration – 14– Continued 64
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- 66. Illustration – 15- Continued 66
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- 68. Illustration – 16- Continued 68
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- 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
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- 75. Illustration 20 -Continued 75
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