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Mathematics
Session
Logarithms
Session Objectives
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 numberother than one
Logarithms Definition
alog N x
Log of N to the
base a is x
xa Nalog N x
22Example : log 4 2 2 4
Note: log of negatives and zero are not Defined in Reals
Illustrative Example
The number log27 is
(a) Integer (b) Rational
(c) Irrational (d) Prime
Solution:
Log27 is an Irrational number
Why?
As there is no rational
number, 2 to the power of which gives 7
Fundamental laws of logarithms
b b b1) log xy log x log y
b bLet log x A, log y B
A Bb x , b y A B A Bxy b b b
b b blog xy A B log x log y hence proved
b b b
x2) log log x log y
y
yb b3) log x ylog x
b b b bExtension log xyz log x log y log z
Other laws of logarithms0
b4) log 1 0 as b 1 1
b5) log b 1 as b b
ab
a
log x6) log x
log b
Change of base
blog x7) b xblog xLet b y blog x
b blog b log y
b b blog xlog b log y b blog x log y
y x
zy
bb
y8) log x log x
z
Where ‘a’ is any other base
Illustrative Example
23Simplify log 2 2
Solution:
23 2
log 2 2 log 2 23
32
2log 2
3
2 3. log 2 log 2
3 2
Illustrative Example
Solution :
log 7 log 33 73 7 True / False ?
log 73log 7 log 73 33 3
1
log 73 log 733
1
log 7 log 33 77 7
Hence True
Illustrative Example
Solution:
If ax = b, by = c, cz = a, then the value of xyz is
a) 0 b) 1 c) 2 d) 3
xa b xloga logb
logbx
loga
logc logaSimilarly y , z
logb logc
Hence xyz 1
Illustrative Example
Find log tan 0.25
Solution:
log tan 0.25 log tan4
log 1 0
Illustrative Example
Solution:
1 1 1
log ...2.5 2 33 3 3Pr ove that 0.16 4
1 1 1 1 / 3
log ... log2.5 2.52 33 1 1 / 33 30.16 0.16
1 / 3
log2.52 / 30.16 1
2log2.520.4 21
log2.5 20.4
21log10 2
4410
21log10 2
4104
21
log 210 2410 1
44 2
Illustrative Example
Solution:
If log32, log3(2x-5) and log3(2x-7/2) are in arithmetic progression, then find the value of x
2log3(2x-5) = log32 + log3(2x-7/2)
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
Why
Illustrative Example
Solution:
If a2+4b2 = 12ab, then prove that log(a+2b) is equal to
1loga logb 4log2
2
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)
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
1 1 1e 1 .... ....
1! 2! n!
Illustrative Example
Solution:
lnln7e 7 True / False ?
Hence False
log blnln7 ae ln7 as a b
Characteristic and Mantissa
Standard form of decimal
pn m 10 where 1 m 10
3Example 1234.56 1.23456 10 30.001234 1.234 10
p pHence log n log m 10 log m log 10
log n log m plog 10 log m p
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 10p , 1 m < 10
log n p log m
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
n Std. formm x 10p
p m ‘abcd’
1234.56 1.23456x103 3 1.2345 1234
0.000123 1.23x10-4 -4 1.23 1230
100 1x102 2 1 1000
0.10023 1.0023x10-1 -1 1.0023 1002
Example n = m x 10p,
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 10p
2) Significant digits of m: ‘abcd’
3) Find number at (ab,c), say x, where ab: row, c: col4) 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(1234.56)
n Std. formm x 10p
p m ‘abcd’
1234.56
1.23456x103
3 1.2345 1234
1) Std. Form n = 1.23456 x 103
2) Significant digits of m: 1234
3) Number at (12,3) = 08994) Number at (12,4) = 14
5) log(n) = 3 + .(0899+14) = 3 + 0.0913 = 3.0913
Note this
Illustrative Example
Find log(0.000123)
n Std. formm x 10p
p m ‘abcd’
0.000123
1.23x10-4 -4 1.23 1230
1) Std. Form n = 1.23 x 10-4
2) Significant digits of m: 1230
3) Number at (12,3) = 08994) As d = 0, y = 0 Note this
5) log(n) = -4 + .(0899+0) = -4 + 0.0899 = -3.9101
To avoid the
calculations4.0899
Illustrative Example
Find log(100)
n Std. formm x 10p
p m ‘abcd’
100 1x102 2 1 1000
1) Std. Form n = 1 x 102
2) Significant digits of m: 1000
3) Number at (10,0) = 00004) As d = 0, y = 0
5) log(n) = 2 + .(0000+0) = 2 + 0.0000 = 2
Illustrative Example
Find log(0.10023)
n Std. formm x 10p
p m ‘abcd’
0.10023
1.0023x10-
1
-1 1.0023 1002
1) Std. Form n = 1.0023 x 10-1
2) Significant digits of m: 1002
3) Number at (10,0) = 00004) Number at (10,2) = 9
5) log(n) = -1 + .(0000+9) = -1 + 0.0009 = -0.9991
To avoid the
calculations1.0009
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 n m.abcd
For eg. If n = -1.2718 = -1 – 0.2718
n = -1-0.2718=-2+1-0.2718
n = -2+0.7282
2.7282
Now n = m.abcd or n m.abcd
How to find Antilog(n)
2) Step2: Select row ‘ab’
Select the row ‘ab’ from the antilog tableEg. n = -1.2718 2.7282
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
Eg. n 2.7282
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
Eg. n 2.7282
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 10m+1
If i.e. n < 0
Antilog(n) = .(x+y) x 10-(m-1)
n m.abcd
Eg. n 2.7282
x = 5346 y = 2
Antilog(n) = .(5346 + 2) x 10-(2-1)
= .5348 x 10-1 = 0.05348
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 103+1
= 0.1234 x 104
= 1234
Solution:
Illustrative Example
Find Antilog(-3.9101)
1) Std. Form n = -3.9101
2) Row 08
3) Number at (08,9) = 12274) Number at (08,9) = 3
5) Antilog(-3.9101)
Solution:
n = -3 – 0.9101 = -4 + 1 – 0.9101n = -4 + 0.0899 4.0899 m.abcd
Antilog 4.0899
= .(1277+3) x 10-(4-1)
= 0.1280 x 10-3
= 0.0001280
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 102+1
= 0.1000 x 103
= 100
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 1.0009
Antilog 1.0009
= .(1000+2) x 10-(1-1)
= 0.1002
Applications
1) Use in Numerical Calculations
2) Calculation of Compound Interest
3) Calculation of Population Growth
4) Calculation of Depreciation
nr
A P 1100
Now take log
n
n o
rp p 1
100Now take log
t
t o
rv v 1
100Now take log
Illustrative Example
Find 3563.4 0.4573
36.15
Solution:
3
3
563.4 0.4573let x
6.15
3
3
563.4 0.4573logx log
6.15
33log 563.4 0.4573 log 6.15
1
log 563.4 log 0.4573 3log 6.153
Solution Cont.
1
log 563.4 log 0.4573 3log 6.153
2 11log 5.634 10 log 4.573 10 3log 6.15
3
1 1
log 5.634 2 log 4.573 3log 6.153 3
1 1
.7508 2 0.6602 3 0.78893 3
= 0.2708
x = antilog (0.2708) = 0.1865 × 101
= 1.865
Illustrative Example
Solution:
Find the compound interest on Rs. 20,000 for 6 years at 10% per annum compounded annually.
n 6r 10
As A P 1 20000 1100 100
= 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
Solution Cont.
log A = 4.5494
A = antilog (4.5494)
= 0.3543 × 105
= 35430
Compound interest = 35430 – 20000 = 15,430
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.
n
n o
rAs p p 1
100
2
2
7.5p 80000 1
100
= 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
Solution Cont.
log p2 = 4.9659
p2 = antilog (4.9659)
= 0.9245 × 105
= 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.
t
t o
rAs v v 1
100
3
2
2v 6250 1
100
= 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)
Solution Cont.
log v2 = 0.7959 + 3 × (0.9912)
= 3.7695
v2 = antilog (3.7695)
= 0.5882 × 104
= Rs. 5882
Class Exercise
Class Exercise - 1
Find 2 3
log 1728
Solution :
2 3
log 1728 x
x 6
6 3 62 3 1728 2 .3 2 . 3
x 6
2 3 2 3
x 6
Class Exercise - 2
Solution :
If a2 + b2 = 7ab, prove that
1
log a b log3 loga logb2
a2 + b2 = 7ab
a2 + b2 + 2ab = 9ab (a + b)2 = 9ab
a b 3 ab
1
2a b
ab3
taking log both sides we get
1
2a blog log ab
3
1log a b log3 logab
2
1
loga logb2
Class Exercise - 3
Solution :
Find x, y if logx log36 log64
log5 log6 logy
2logx log36 log6 2log6
2log5 log6 log6 log6
logx = 2 log5 = log52 = log25
x = 25Similarly
1
2log64 1
2 logy log64 log64 log8logy 2
y = 8
Class Exercise - 4
Solution :
If 210 100log x log y 1 find y if x = 2.
210 100log x log y 1
2 1010
10
log ylog x 1
log 100 2 10
10
log ylog x 1
2
210 10
1log x log y 1
2
11 2
2 210 10log x log y 1
2
10 1
4
xlog 1
y
2
1
4
x10
y
4 42x 4 16y
10 10 625
Class Exercise - 5
Solution :
Simplify
5log x 2log 8b 71 3
2 4b and 7
(i)
55log xb1 5log x 2b log xb2 2b b b
5
2x
(ii)
1
22log 873
47
3 1/ 2log 8727
31 22log 87
7
31 322 48 8
Class Exercise - 6
Solution :
Simplify x 2 3x xlog 4 log 16 log 64 12
and x = 2k then k is
(a) 0.25 (b) 0.5
(c) 1 (d) 2
x 2 3x xlog 4 log 16 log 64 12
2 3
log4 log16 log6412
logx logx logx
log4 log16 log6412
logx 2logx 3logx
11
32log4 log16 log6412
logx logx logx
log4 log4 log412
logx logx logx
3log4 3
12 logx log4logx 12
Class Exercise – 7 (i)
Solution :
(i) If x, y, z > 0, such that
logx logy logz,
y z z x x yevaluate xx yy zz.
logx logy logzsay
y z z x x y
logx y z , logy z x , logz x y
x logx xy xz y logy yz xy z logz zx yz
x logx + y logy + z logz xy xz yz xy zx yz 0
x y zlogx logy logz 0
x y zlogx .y .z 0 xx.yy.zz = 1
Class Exercise – 7 (ii)
a b c a b c a b c a b c 1
sayloga logb logc
loga a b c a , logb b c a b , logc c a b c
bloga alogb ab b c a ab c a b
ab b c a c a b 2 abc
(ii) If a, b, c > 0, such that
a b c a b c a b c a b c
loga logb logc
then prove that ab ba = bc cb = ca ac
Solution :
Solution Cont.
bloga alogb 2 abc
Similarly c logb blogc 2 abc and
alogc c loga 2 abc
Hence b loga + a logb = c logb + b logc= a logc + c loga
logab.ba = logbc cb = logca ac
b a c b a ca b b c c a
Class Exercise - 8
Solution :
Find characteristic, mantissa and log of each of the following
(i) 67.77 (ii) .0087
(i) 67.77 = 6.777 × 101
Characteristic = 1 Mantissa = log (6.777)
= 0.(8306+5)
= 0.8311
log 67.77 = 1 + 0.8311 = 1.8311
Solution Cont.
(ii) 0.087 = 8.7 × 10–3
Characteristic = –3
Mantissa = log (8.7) = 0.(9395 + 0)
= 0.9395
log (0.008) = -3 + 0.9395 = 3.9395
Class Exercise 9
Solution
Find the antilogarithm of each of the following
(i) 4.5851 (ii) –0.7214
(i) Antilog(4.5851)
= .(3846 + 1) × 105
= 38470
(ii) Antilog(–0.7214) = Antilog(–1 + 1 – 0.7214)
= .(1897 + 3) × 100
= 0.19
Antilog(–1 + 0.2786) = Antilog 1.2786
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.
nr
A P 1100
313125
100900 P 1 P 1.25100
31
100900P
1.25logP = log(100900) – 31log (1.25)
= log (1.009 × 105) – 31log (1.25)
= log (1.009) + 5 – 31 log (1.25)
Solution Cont.
log P = 1.9998
P = Antilog (1.9998)
= 0.9995 × 102
= 99.95
= 0.0037 + 5 – 31 (0.0969)
= 5.0037 – 3.0039
= 1.9998
Thank you