Transcript Document

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
Logarithms Definition
Base: Any postive real number
other than one
loga N  x
Log of N
to the
base a is x
loga N  x
 ax  N
Note: log of negatives and
Example : log2 4  2  22  4 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
1) logb xy  logb x  logb y
Let logb x  A, logb y  B
 b A  x , bB  y
 xy  bAbB  bA B
 logb xy  A  B  logb x  logb y hence proved
Extension logb xyz  logb x  logb y  logb z
2) logb
x
 logb x  logb y
y
3) logb xy  y logb x
Other laws of logarithms
4) logb 1  0 as b0  1
5) logb b  1 as b1  b
loga x
6) logb x 
loga b
Change
of base
Where ‘a’ is any other base
7) blogb x  x
log x
Let blogb x  y  logb b b  logb y
 logb xlogb b  logb y  logb x  logb y
yx
y
8) logbz x  logb x
z
y
Illustrative Example

Simplify log 2 2

2
3
Solution:

log 2 2

2
3


2
log 2 2
3

 23 
2

log  2 
3
 

2 3
. log 2   log 2
3 2
Illustrative Example
3
log3 7
7
log7 3
Solution :
3
log3 7

 3
3
log3 7

log3 7
log3 7
1
log3 7
1
7
log3 7
7
Hence True
log7 3
True / False ?
Illustrative Example
If ax = b, by = c, cz = a, then the
value of xyz is
a) 0
b) 1
c) 2
d) 3
Solution:
ax  b  xloga  logb
x
logb
log a
Similarly
logc
loga
y
, z
logb
logc
Hence xyz  1
Illustrative Example
Find log tan 0.25 
Solution:
 
log tan 0.25   log tan  
 4
 log 1  0
Illustrative Example
1 1 1

log2.5  
 ...
 3 32 33

0.16
Pr ove that
 4
Solution:
0.16
1 1 1

log2.5  

...
 3 32 33

 0.16 
1/3
log2.5
2 /3
 0.16
 4
 
 10 
 10 
 
 4
 1
log10  
 2
4
 1
log10  
 2
4
2
1/3
log2.5
1 1 / 3
 0.4
 10 
 
 4
2
 1
 
 2
1
2log2.5
2
 1
 log10  
 2
4
2
4
2
 1
log2.5  
 2
 0.4
2
Illustrative Example
If log32, log3(2x-5) and log3(2x-7/2)
are in arithmetic progression, then
find the value of x
Solution:
2log3(2x-5) = log32 + log3(2x-7/2)
log3(2x-5)2 = log32.(2x-7/2)
Why
(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
Illustrative Example
If a2+4b2 = 12ab, then prove that
log(a+2b) is equal to
1
loga  logb  4log2

2
Solution:
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
e 1
1 1
1

 .... 
 ....
1! 2!
n!
Illustrative Example
elnln7  7 True / False ?
Solution:
elnln7  ln7 as aloga b  b
Hence False
Characteristic and
Mantissa
Standard form of decimal
n  m  10p where 1  m  10
Example 1234.56  1.23456  103
0.001234  1.234  103



Hence log n  log m  10p  log m  log 10p

log n  log m  plog 10   log m  p
p is characteristic of n log(n)=mantissa+characteristic
log(m) is mantissa of n
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
Example
n = m x 10p,
p: characteristic, log(m): mantissa
n
Std. form
m 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
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
What if d = 0?
‘ab’, let it be y
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:
Never neglect 0’s
at end or front
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)
Illustrative Example
Find log(1234.56)
n
Std. form
m x 10p
p
m
‘abcd’
1234.5
6
1.23456x1
03
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
Note this
5) log(n) = 3 + .(0899+14) = 3 + 0.0913 = 3.0913
Illustrative Example
Find log(0.000123)
n
Std. form
m x 10p
p
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
To avoid
the
calculations
4.0899
5) log(n) = -4 + .(0899+0) = -4 + 0.0899 = -3.9101
Illustrative Example
Find log(100)
n
Std. form
m 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) = 0000
4) As d = 0, y = 0
5) log(n) = 2 + .(0000+0) = 2 + 0.0000 = 2
Illustrative Example
Find log(0.10023)
n
Std. form
m x 10p
p
m
‘abcd’
0.1002
3
1.0023x10
-1
1.0023
1002
-1
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
To avoid
the
calculations
1.0009
5) log(n) = -1 + .(0000+9) = -1 + 0.0009 = -0.9991
How to find Antilog(n)
(1) Step1: Standard form of number
If n  0, say n = m.abcd
If n < 0, convert it into bar
notation say n  m.abcd
For eg. If n = -1.2718 = -1 – 0.2718
For bar notation subtract 1, add 1 we get
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 table
Eg. 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 n  m.abcd i.e. n < 0
Antilog(n) = .(x+y) x 10-(m-1)
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)
Solution:
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
Illustrative Example
Find Antilog(-3.9101)
Solution:
1) Std. Form n = -3.9101
n = -3 – 0.9101 = -4 + 1 – 0.9101
n = -4 + 0.0899  4.0899  m.abcd
2) Row 08
3) Number at (08,9) = 1227
4) Number at (08,9) = 3
5) Antilog(-3.9101)  Antilog 4.0899

= .(1277+3) x 10-(4-1)
= 0.1280 x 10-3
= 0.0001280

Illustrative Example
Find Antilog (2)
Solution:
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)
= .(1000+0) x 102+1
= 0.1000 x 103
= 100
Illustrative Example
Find Antilog(-0.9991)
Solution:
1) Std. Form n = -0.9991
-0.9991 = -1 + 1 – 0.9991
= -1 + 0.0009  1.0009
2) Row 00
3) Number at (00,0) = 1000
4) Number at (00,9) = 2

5) Antilog(-0.9991)  Antilog 1.0009
= .(1000+2) x 10-(1-1)
= 0.1002

Applications
1) Use in Numerical Calculations
2) Calculation of Compound Interest
r 

A  P 1 
 100 
n
Now take log
3) Calculation of Population Growth
r 

pn  po  1 
 100 
n
Now take log
4) Calculation of Depreciation
r 

v t  v o 1 
 100 
t
Now take log
Illustrative Example
Find
563.4  3 0.4573
6.15 3


Solution:
let x 
563.4  3 0.4573
log x  log
 6.15
3
563.4  3 0.4573
 6.15
 log  563.4  0.4573   log  6.15

3
3
3


1
 log  563.4  log 0.4573  3log  6.15
3
Solution Cont.
1
 log  563.4  log 0.4573  3log  6.15
3




1
 log 5.634  10  log 4.573  101  3log  6.15
3
1
1
 log 5.634  2  log  4.573   3log  6.15
3
3
1
1
 .7508  2  0.6602   3 0.7889 = 0.2708
3
3
2
x = antilog (0.2708) = 0.1865 × 101
= 1.865
Illustrative Example
Find the compound interest on Rs.
20,000 for 6 years at 10% per
annum compounded annually.
Solution:
n
r 
10 


As A  P  1 
 20000  1 

 100 
 100 
= 20000 (1.1)6
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
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
r


Solution: As pn  po  1 
 100 
2
7.5 

p2  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
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
r

Solution: As v  v  1 
t
o 
 100 
2 

v 2  6250  1 
 100 
3
= 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 log2
3
1728
Solution :
log2 3 1728  x
   1728  2 .3
 2 3   2 3 
 2 3
x
x
 x6
6
6
3
2.
6
 3
6
Class Exercise - 2
If a2 + b2 = 7ab, prove that
1
log  a  b  log3  loga  logb 
2
Solution :
a2 + b2 = 7ab
a2 + b2 + 2ab = 9ab
(a + b)2 = 9ab
 a  b  ab 21
a  b  3 ab
 
3
taking log both sides we get
a  b

log
 log
3

ab
1
2
1
loga  logb
2
1
log  a  b  log3  logab
2
Class Exercise - 3
log x log36 log64


Find x, y if
log5 log6
log y
Solution :
log x log36 log62 2log6



2
log5 log6
log6
log6
logx = 2 log5 = log52 = log25
Similarly x = 25
1
log64
1
 2  log y  log64  log64 2  log8
log y
2
y=8
Class Exercise - 4
If log10 x 2  log100 y  1 find y if x = 2.
Solution :
log10 x 2  log100 y  1
log10 x 2 
log10 y
log10 y
 1 log10 x 2 
1
log10 100
2
1
2
1
 
log10 x  log10 y  1 log10 x 2  log10  y   1
2
 
1
2
2
log10
x2
y
1
4
1
x2
y
4
1
4
 10
 x2 
16
 4
y    
 10 
625
 10 
4
Class Exercise - 5
 
 b 
1
2
Simplify
5logb x
 
and  7 
 
3
4
2log7 8
Solution :
(i)
 
 b 
1
2
 34 
(ii)  7 
 
7
5logb x
b
 1
2log7  8 2 
 
 
3
 1 2
log7  8 2 
 
 
x
5
log x
2 b
b
5
logb x 2
5
2
7
 
 8 
 
1
2
 
3
 log7 81/ 2
2
3
2
8
3
4
Class Exercise - 6
Simplify logx 4  log 2 16  log 3 64  12
x
x
and x = 2k then k is
(a) 0.25
(b) 0.5
(c) 1
(d) 2
Solution :
logx 4  log 2 16  log 3 64  12
x
x
log 4 log16 log64


 12
2
3
log x log x
log x
1
2
1
3
log 4 log16
log64


 12
log x
log x
log x
3log 4
3
 12  log x 
log 4
log x
12
log 4 log16 log64


 12
log x 2log x 3log x
log 4 log 4 log 4


 12
log x log x log x
Class Exercise – 7 (i)
(i) If x, y, z > 0, such that
log x log y log z


, evaluate xx yy zz.
yz zx xy
Solution :
log x log y log z



yz zx xy
 say 
log x    y  z  , log y    z  x  , log z    x  y 
x log x    xy  xz 
y log y    yz  xy 
z log z    zx  yz 
x logx + y logy + z logz    xy  xz  yz  xy  zx  yz   0
logx x  logy y  logzz  0
logx x .y y .zz  0
xx.yy.zz = 1
Class Exercise – 7 (ii)
(ii) If a, b, c > 0, such that
a b  c  a
loga

b c  a  b
logb

c a  b  c 
logc
then prove that ab ba = bc cb = ca ac
Solution :
a b  c  a
loga

b  c  a  b
logb

c a  b  c 
logc
1


 say
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
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
 ab ba  bc cb  caac
Class Exercise - 8
Find characteristic, mantissa and log of
each of the following
(i) 67.77
(ii) .0087
Solution :
(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
Find the antilogarithm of each
of the following
(i) 4.5851
(ii) –0.7214
Solution
(i) Antilog(4.5851)
= .(3846 + 1) × 105
= 38470
(ii) Antilog(–0.7214) = Antilog(–1 + 1 – 0.7214)


Antilog(–1 + 0.2786) = Antilog 1.2786 = .(1897 + 3) × 100
= 0.19
Class Exercise - 10
If a sum of money amounts to Rs.
100900 in 31 years at 25% per annum
compound interest, find the sum.
Solution
r 

A  P 1 
 100 
n
25 

100900  P  1 
 100 
P
100900
1.25
31
 P 1.25 
31
logP = log(100900) – 31log (1.25)
31
= log (1.009 × 105) – 31log (1.25)
= log (1.009) + 5 – 31 log (1.25)
Solution Cont.
= 0.0037 + 5 – 31 (0.0969)
= 5.0037 – 3.0039
= 1.9998
log P = 1.9998
P = Antilog (1.9998)
= 0.9995 × 102
= 99.95
Thank you