Mathematics. Session Logarithms Session Objectives.

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


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


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


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


Find log tan 0.25

Solution:

log tan 0.25 log tan4

log 1 0


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


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


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!


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’


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)


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


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


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


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


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


Find log(100)

n Std. formm x 10p

p m ‘abcd’

100 1x102 2 1 1000

1) Std. Form n = 1 x 102


3) Number at (10,0) = 00004) As d = 0, y = 0

5) log(n) = 2 + .(0000+0) = 2 + 0.0000 = 2


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



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


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


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


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


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:


Find Antilog(-3.9101)

1) Std. Form n = -3.9101

2) Row 08


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


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


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


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


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


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


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

Mathematics. Session Logarithms Session Objectives.

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Transcript of Mathematics. Session Logarithms Session Objectives.