introduction to logarithm

Ratio and Proportion, Indices and Logarithm– Chapter 1 Paper 4: Quantitative Aptitude-Statistics

Ms. Ritu Gupta B.A. (Hons.) Maths and MA (Maths)

Introduction to Logarithm

• Fundamental Knowledge • Its application

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Definition of Logarithm

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Example

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Properties of Logarithm

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Things to Remember

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Laws of Logarithm

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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

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Logarithm of a Product Rule Contd…

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

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Logarithm of a Quotient Contd…

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

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Logarithm of a power of a number Contd…

Change of Base

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Change of Base Contd…

Base Changing Result

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Systems of Logarithm

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Common Logarithms

Natural Logarithms

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Systems of Logarithms

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

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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.

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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

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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

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Example – Continued

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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.

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Example

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Rules to find Characterstic

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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

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Number Characteristic 48 1 3578 3 8.31 0

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

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Number Characteristic .6 -1

.09 -2

.00657 -3

.000852 -4

Mantissa • The Mantissa of the common logarithm of a number can

be found from a log-table.

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What is Log Table

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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.

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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

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Example- Continued

Mantissa = 0.(0969+ 21) = 0.0990 log 125.6 = 2 + 0.0990 = 2.0990

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Point to remember

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Point to remember- Continued

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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

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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

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

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Illustrations

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Illustration 1

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Illustration 2

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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

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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)}]

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Illustration 3 – Continued = log2 [log2 {log3 (9X1)}] (as log3 3 = 1)

= log2 [log2 {log3 32}]

= log2 [log2 (2log3 3)]

= log2 [log2 2] = log21 = 0

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Illustration 4

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Illustration 4 – Continued

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Illustration – 5

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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.

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Illustration – 6

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Illustration – 6 – Continued

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Illustration – 7

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Illustration – 7 – Continued

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Illustration – 8

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Illustration – 9

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Illustration – 10

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Illustration – 10 – Continued

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Illustration – 11

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Illustration – 11 – Continued

loga + logb + logc = ky – kz+ kz – kx + kx – ky

log(abc) = 0

log(abc) = log1

abc = 1

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Illustration – 12

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Illustration – 12 – Continued

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Illustration – 13

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Illustration – 13 – Continued

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Illustration – 14

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Illustration – 14 – Continued

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Illustration – 15

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Illustration – 15 - Continued

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Illustration – 16

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Illustration – 16 - Continued

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Illustration – 17

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Illustration – 17 – Continued

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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

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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

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∴

Illustration 19 – Continued

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Illustration 20

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Illustration 20 - Continued

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Thank You!

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