Download - 11. Logarithms

8/10/2019 11. Logarithms

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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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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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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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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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Find log tan 0.25

Solution:

log tan 0.25 log tan4

log 1 0


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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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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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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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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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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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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


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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Find log(100)

n Std. form

m x 10pp m abcd

100 1x102 2 1 1000

1) Std. Form n = 1 x 102


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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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


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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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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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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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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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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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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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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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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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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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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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