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Session
Logarithms
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Session Objectives
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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
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Base:Any postive real number
other than one
Logarithms Definition
alog N x
Log of Nto the
base a is x
xa Nalog N x
2
2Example : log 4 2 2 4 Note: log of negatives andzero are not Defined in Reals
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Illustrative Example
The number log27 is
(a) Integer (b) Rational
(c) Irrational (d) Prime
Solution:
Log27 is an Irrational number
Why?
As there is norational number,
2 to the powerof which gives 7
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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 bx2) log log x log yy
y
b b3) log x y log x
b b b bExtension log xyz log x log y log z
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Other laws of logarithms
0
b4) log 1 0 as b 1
1
b5) log b 1 as b b
ab
a
log x6) log x
log b
Changeof base
blog x7) b x
blog xLet b y blog x
b blog b log y
b b blog xlog b log y b blog x log y
y x
z
y
bb
y8) log x log x
z
Where a is any other base
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Illustrative Example
2
3
Simplify log 2 2
Solution:
2
3 2log 2 2 log 2 23
3
22
log 23
2 3
. log 2 log 23 2
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Illustrative Example
Solution :
log 7 log 33 7
3 7 True / False ?
log 73
log 7 log 73 33 3
1
log 73 log 733
1
log 7 log 33 7
7 7
Hence True
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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 , zlogb logc
Hence xyz 1
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Illustrative Example
Find log tan 0.25
Solution:
log tan 0.25 log tan4
log 1 0
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Illustrative Example
Solution:
1 1 1log ...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
2log2.5 20.4 21
log2.5 20.4
21
log10 24
4
10
21
log10 24
10
4
2
1log 210 2
410 1
44 2
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Illustrative Example
Solution:
If log32, log3(2x-5) and log3(2
x-7/2)
are in arithmetic progression, thenfind the value of x
2log3(2x-5) = log32 + log3(2
x-7/2)
log3(2x-5)2= log32.(2
x-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
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Illustrative Example
Solution:
If a2+4b2= 12ab, then prove that
log(a+2b) is equal to
1
loga logb 4log22
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) = (4log 2 + log a + log b)
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System of logarithms
Common logarithm:Base = 10
Log10x, also known as Briggssystem
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!
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Illustrative Example
Solution:
lnln7e 7 True / False ?
Hence False
log blnln7 ae ln7 as a b
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Characteristic andMantissa
Standard form of decimal
pn m 10 where 1 m 10
3Example 1234.56 1.23456 10
3
0.001234 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 characteristicof n
log(m) is mantissaof n
log(n)=mantissa+characteristic
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How to find log(n) using logtables
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 tofind the mantissa of n i.e. log(m)
2) Step2: Significant digits
Identify 4 digits from left, starting from first nonzero digit of m, inserting zeros at the end ifrequired, let it be abcd
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How to find log(n) using logtables
n Std. form
m x 10pp 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)
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How to find log(n) using log tables
3) Step3: Select row ab
Select row ab from thelogarithmic table
4) Step4: Select column c
Locate number at column cfrom the row ab, let it be x
5) Step5: Select column of mean difference d
If d 0,Locate number at column dof mean difference from the rowab, let it be y
What if d = 0?Consider y = 0
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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: col
4) Find number at (ab,d), say y, where d: mean diff
5) log(n) = p + .(x+y)
Never neglect 0s
at end or front
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Illustrative Example
Find log(1234.56)
n Std. form
m x 10pp 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) = 0899
4) Number at (12,4) = 14
5) log(n) = 3 + .(0899+14) = 3 + 0.0913 = 3.0913
Note this
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Illustrative Example
Find log(0.000123)
n Std. form
m x 10pp m abcd
0.0001
23
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) = 0899
4) As d = 0, y = 0 Note this
5) log(n) = -4 + .(0899+0) = -4 + 0.0899 = -3.9101
To avoidthe
calculations
4.0899
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Illustrative Example
Find log(100)
n Std. form
m x 10pp 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) = 0000
4) As d = 0, y = 0
5) log(n) = 2 + .(0000+0) = 2 + 0.0000 = 2
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Illustrative Example
Find log(0.10023)
n Std. form
m x 10pp 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) = 0000
4) Number at (10,2) = 9
5) log(n) = -1 + .(0000+9) = -1 + 0.0009 = -0.9991
To avoidthe
calculations
1.0009
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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 barnotation 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
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How to find Antilog(n)
2) Step2: Select row ab
Select the row ab fromthe antilog table
Eg. n = -1.2718 2.7282
Select row 72 from table
3) Step3: Select column c of ab
Select the column c ofrow ab from the antilogtable, locate the number
there, let it be x
Eg. n 2.7282
Number at col 8 of row72 is 5346, x = 5346
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How to find Antilog(n)
4) Step4: Select col. d of mean diff.Select the col d of meandifference of the row abfrom the antilog table, letthe number there be y, Ifd = 0, take y as 0
Eg. n 2.7282
Number at col 2 of meandiff. of row 72 is 2, y = 2
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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
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Illustrative Example
Find Antilog(-3.9101)
1) Std. Form n = -3.9101
2) Row 08
3) Number at (08,9) = 1227
4) Number at (08,9) = 3
5) Antilog(-3.9101)
Solution:
n = -3 0.9101 = -4 + 1 0.9101
n = -4 + 0.0899 4.0899 m.abcd
Antilog 4.0899= .(1277+3) x 10-(4-1)
= 0.1280 x 10-3
= 0.0001280
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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
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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
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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
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Illustrative Example
Find
3563.4 0.4573
36.15
Solution:
3
3
563.4 0.4573
let x 6.15
3
3
563.4 0.4573logx log
6.15
3
3log 563.4 0.4573 log 6.15
1
log 563.4 log 0.4573 3log 6.153
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Solution Cont.
1log 563.4 log 0.4573 3log 6.153
2 11
log 5.634 10 log 4.573 10 3 log 6.153
1 1log 5.634 2 log 4.573 3 log 6.153 3
1 1
.7508 2 0.6602 3 0.78893 3
= 0.2708
x = antilog (0.2708) = 0.1865 101
= 1.865
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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
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Solution Cont.
log A = 4.5494
A = antilog (4.5494)
= 0.3543 105
= 35430
Compound interest = 35430 20000 = 15,430
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Illustrative Example
Solution:
The population of the city is 80000. If the
population increases annually at the rateof 7.5%, find the population of the city
after 2 years.
n
n o
rAs p p 1
100
2
27.5p 80000 1100
= 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
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Solution Cont.
log p2= 4.9659
p2= antilog (4.9659)
= 0.9245 105
= 92450
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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 (9.8 101)
= log 6.250 + 3 + 3 log (9.8) 3
= 0.7959 + 3 (0.9912)
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Solution Cont.
log v2= 0.7959 + 3 (0.9912)
= 3.7695v2= antilog (3.7695)
= 0.5882 104
= Rs. 5882
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Class Exercise
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Class Exercise - 1
Find2 3
log 1728
Solution :
2 3
log 1728 x
x 66 3 62 3 1728 2 .3 2 . 3
x 6
2 3 2 3
x 6
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Class Exercise - 2
Solution :
If a2+ b2= 7ab, prove that
1log 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 logb
2
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Class Exercise - 3
Solution :
Find x, y if log x log 36 log 64
log5 log 6 log y
2logx log36 log6 2 log6
2log5 log6 log6 log6
logx = 2 log5 = log52= log25
x = 25Similarly
12log64 12 logy log64 log64 log8
logy 2
y = 8
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Class Exercise - 4
Solution :
If 210 100
log x log y 1 find y if x = 2.
210 100
log x log y 1
2 1010
10
log ylog x 1
log 100
2 1010
log ylog x 1
2
210 10
1log x log y 1
2
11 2
2 210 10
log x log y 1
2
10 1
4
xlog 1
y
2
1
4
x
10y
4 42x 4 16y
10 10 625
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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
22 log 873
47 3 1/2log 8727
31 22log 87
7
31 322 48 8
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Class Exercise - 6
Solution :
Simplify x 2 3x x
log 4 log 16 log 64 12
and x = 2kthen k is
(a) 0.25 (b) 0.5
(c) 1 (d) 2
x 2 3x x
log 4 log 16 log 64 12
2 3
log4 log16 log6412
logx logx logx
log4 log16 log6412
logx 2 logx 3 logx
11
32log4 log16 log6412
logx logx logx log4 log4 log4 12
logx logx logx
3log4 3
12 logx log4logx 12
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Class Exercise 7 (ii)
a b c a b c a b c a b c 1say
loga 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 cloga logb logc
then prove that abba= bccb= caac
Solution :
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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= logbccb= logcaac
b a c b a ca b b c c a
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Class Exercise - 8
Solution :
Find characteristic, mantissa and log ofeach 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
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Solution Cont.
(ii) 0.087 = 8.7 103
Characteristic = 3
Mantissa = log (8.7) = 0.(9395 + 0)
= 0.9395
log (0.008) = -3 + 0.9395 = 3.9395
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Class Exercise 9
Solution
Find the antilogarithm of eachof 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
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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 1
100
31
3125100900 P 1 P 1.25
100
31
100900P
1.25
logP = log(100900) 31log (1.25)
= log (1.009 105) 31log (1.25)
= log (1.009) + 5 31 log (1.25)
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Thank you