Mathematics

Session Logarithms

Session Objectives

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

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

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

Fundamental laws of logarithms

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

Illustrative Example Solution:

Illustrative Example Solution : True / False ? Hence True

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

Illustrative Example Solution:

Illustrative Example Solution:

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 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 x + 32 = 0, put 2 x = y, we get y y + 32 = 0  (y-4)(y-8) = 0  y = 4 or 8  2 x =4 or 8  x = 2 or 3 Why

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)

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

Illustrative Example Solution: True / False ? Hence False

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

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’

How to find log(n) using log tables nStd. form m x 10 p pm‘abcd’ x x x x Example n = m x 10 p, p: characteristic, log(m): mantissa Log(n) = p + log(m)

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

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

Illustrative Example Find log( ) nStd. form m x 10 p pm‘abcd’ x ) Std. Form n = x ) Significant digits of m: ) Number at (12,3) = ) Number at (12,4) = 14 5) log(n) = 3 +.( ) = = Note this

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

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

Illustrative Example Find log( ) nStd. form m x 10 p pm‘abcd’ x ) Std. Form n = x ) Significant digits of m: ) Number at (10,0) = ) Number at (10,2) = 9 5) log(n) = -1 +.(0000+9) = = To avoid the calculations

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 – n = = n = Now n = m.abcd or

How to find Antilog(n) 2) Step2: Select row ‘ab’ Select the row ‘ab’ from the antilog table Eg. n = 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

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

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) =.( ) x 10 -(2-1) =.5348 x =

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

Illustrative Example Find Antilog( ) 1) Std. Form n = ) Row 08 3) Number at (08,9) = ) Number at (08,9) = 3 5) Antilog( ) Solution: n = -3 – = – n = =.(1277+3) x 10 -(4-1) = x =

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

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

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

Illustrative Example Find Solution:

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

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

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

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

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

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 log 0.98 = log (6.250 × 10 3 ) + 3 log (9.8 × 10 –1 ) = log log (9.8) – 3 = × (0.9912)

Solution Cont. log v 2 = × (0.9912) = v 2 = antilog (3.7695) = × 10 4 = Rs. 5882

Class Exercise

Class Exercise - 1 Find Solution :

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

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

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

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

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

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

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 :

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

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

Solution Cont. (ii) = 8.7 × 10 –3 Characteristic = –3 Mantissa = log (8.7) = 0.( ) =

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

Class Exercise - 10 Solution If a sum of money amounts to Rs 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)

Solution Cont. log P = P = Antilog (1.9998) = × 10 2 = = – 31 (0.0969) = – =

Thank you