Std 11 Commerce, Mathematics and Statistics - II, · PDF file ·...

Std. XI Commerce

Mathematics & Statistics - II

Written as per the revised syllabus prescribed by the Maharashtra State Board

of Secondary and Higher Secondary Education, Pune.

Printed at: Repro India Ltd. Mumbai

P.O. No. 24855

10163_10830_JUP

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Salient Features

• Exhaustive coverage of entire syllabus.

• Topic-wise distribution of all textual questions and practice problems at the

beginning of every chapter.

• Covers answers to all textual and miscellaneous exercises.

• Precise theory for every topic.

• Neat, labelled and authentic diagrams.

• Relevant and important formulae wherever required.

Preface Mathematics is not just a subject that is restricted to the four walls of a classroom. Its philosophy and applications are to be looked for in the daily course of our life. The knowledge of mathematics is essential for us, to explore and practice in a variety of fields like business administration, banking, stock exchange and in science and engineering.

With the same thought in mind, we present to you “Std. XI Commerce: Mathematics and Statistics-II” a complete and thorough book with a revolutionary fresh approach towards content and thus laying a platform for an in depth understanding of the subject. This book has been written according to the revised syllabus.

At the beginning of every chapter, topic–wise distribution of all textual questions including practice problems have been provided for simpler understanding of different types of questions. Neatly labelled diagrams have been provided wherever required. We have provided answer keys for all the textual questions and miscellaneous exercises. In addition to this, we have included practice problems based upon solved exercises which not only aid students in self evaluation but also provide them with plenty of practice. We’ve also ensured that each chapter ends with a set of Multiple Choice Questions so as to prepare students for competitive examinations.

We are sure this study material will turn out to be a powerful resource for students and facilitate them in understanding the concepts of Mathematics in the most simple way.

The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you.

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Best of luck to all the aspirants!

Yours faithfully Publisher

No. Topic Name Page No.

1 Logarithms 1 2 Theory of Attributes 29 3 Partition Values 58 4 Measures of Dispersion 125 5 Moments 162 6 Skewness and Kurtosis 186 7 Permutations and Combinations 222 8 Probability 263 9 Index Numbers 293 10 Time Series 327

1

Chapter 01: Logarithms

Type of Problems Exercise Q. Nos.

Problems based on definition of

logarithm

1.1

Q.1 (i. to iv.)

Q.2 (i. to iv.)

Q.3 (i. to iv.)

Practice Problems

(Based on Exercise 1.1)

Q.1 (i. to iv.)

Q.2 (i. to iv.)

Q.3 (i. to iv.)

Miscellaneous Q.12

Practice Problems

(Based on Miscellaneous) Q.4, 8

Law of Product

1.1

Q.7 (i.)

Q.8 (ii.)

Q.11 (i.)

Practice Problems

(Based on Exercise 1.1) Q.6 (i.)

Miscellaneous Q.1, 3, 4, 17

Practice Problems

(Based on Miscellaneous) Q.2

Law of Exponent

1.1 Q.5 (i.)

Miscellaneous Q.2

Practice Problems


Law of Quotient 1.1 Q.7 (ii.)

Miscellaneous Q.9

Problems based on Product,

Quotient and Exponent laws 1.1

Q.4 (i. to v.)

Q.5 (ii to v.)

Q.6 (i. to iii.)

Q.7 (iii., iv.)

Logarithms01

2

Std. XI : Commerce (Maths ‐ II)

Q.8 (i., iii.)

Q.9 (i. to iii.)

Q.11 (ii., iii.)

Q.12 to Q.14

Practice Problems

(Based on Exercise 1.1)

Q.4, 5

Q.6 (ii.)

Q.7, 8, 10, 11

Miscellaneous Q.5 to Q.8

Q.10, 11, 14, 16, 19, 20

Practice Problems

(Based on Miscellaneous) Q.3, 5, 6, 7, 12, 15

To solve problems without using

log table

1.1 Q.10 (i. to iv.)

Practice Problems

(Based on Exercise 1.1) Q.9 (i., ii.)

Miscellaneous Q.13

Practice Problems


Change of Base law

1.2 Q.1 to Q.12

Practice Problems

(Based on Exercise 1.2) Q.1 to Q.7

Miscellaneous Q.15, 18

Practice Problems

(Based on Miscellaneous) Q.10, 11, 13, 14

To solve problems by using log

table

1.3 Q.1 to Q.12

Practice Problems

(Based on Exercise 1.3) Q.1 to Q.7

Miscellaneous Q.21 to Q.24

Practice Problems


3


Syllabus: 1.1 Definition 1.2 Laws of Logarithms 1.3 Change of base law 1.4 Numerical problems Introduction In mathematics, logarithm of a number to a given base is the power of exponent to which the base must be raised in order to produce the number.

For example, the logarithm of 32 to the base 2 is 5 because 5 is how many 2s one must multiply to get 32. Thus 2 2 2 2 2 = 32.

In the language of exponent, 25 = 32 so log2 32 = 5. 1.1 Definition If ax = b, then x = logab (a > 0, a ≠ 1), (b > 0)

where a is called the base of the logarithm. The two statements ax = b and x = logab are equivalent.

The statement ax = b is said to be in the exponential form and the statement x = logab is said to be in the logarithmic form.

We can convert an exponential form into the logarithmic form. Example:

Exponential form Logarithmic form

34 = 81 log381= 4

25 = 32 log232 = 5

32 = 1

9

3

1log

9

= 2

21 = 2 log22= 1 Remarks

1. We have m = ax if and only if x = loga m 2. Negative numbers and zero have no logarithms.

3. i. loga 1 = 0, a > 0, a 1

Proof:

Let loga 1= x

ax = 1 = a0

x = 0

loga 1 = 0

i.e., logarithm of 1 to any base is 0

ii. loga a = 1, a > 0, a 1 Proof : Let loga a = x ax = a = a1 x = 1 loga a = 1 i.e., logarithm of a number to the same

base is 1.

iii. aloga x x , a > 0, a 1 Proof: Let loga x = y ay = x

aloga x x

iv. If logam = loga n, then m = n.

v. If a > 1 and m > n, then loga m > loga n and conversely.

1.2 Laws of Logarithms 1. Law of Product: loga (xy) = loga x + loga y, (a, x, y > 0, a ≠ 1) Proof: Let loga x = m and loga y = n By definition of logarithm, we get x = am and y = an am. an = xy am + n = xy By definition of logarithm, we get loga (xy) = m + n loga (xy) = loga x + loga y Thus logarithm of the product of two numbers

is equal to the sum of their logarithms with reference to the same base.

Corollary: loga(xyz…) = loga x + loga y + loga z +….. 2. Law of Quotient:

loga

x

y= loga x loga y, (a, x, y > 0, a ≠ 1)

Proof: Let loga x = m and loga y = n By definition of logarithm, we get am = x and an

= y

m

n

a

a =

x

y

am n =

x

y

4


By definition of logarithm, we get

loga

x

y = m n

loga

x

y= loga x loga y

Corollary:

i. loga1

x

= loga 1 loga x = 0 loga x

= loga x

ii. loga a alog logpq

xyx y a alog p log q

3. Law of Exponent: loga x

y = yloga x, (x > 0, a > 0, a ≠ 1) Proof: Let loga x = m By definition of logarithm, we get x = am

Now, xy = may

xy = amy By definition of logarithm, we get loga x

y = my

loga xy = yloga x

Corollary:

i. logam x =

1

mloga x

ii. loga

p q

r sz w

x y= p loaa x

+ q loga y – r loga z

– s loga w

Exercise 1.1 1. Write the following in logarithmic form: i. 83 = 512 ii. 323/5 = 8

iii. 72 = 1

49 iv. 102 = 0.01

Solution: i. 83 = 512

3 = log8 512 ….[By definition of logarithm]

ii. 3

532 = 8

3

5 = log32 8 ….[By definition of logarithm]

iii. 72 = 1

49

– 2 = log71

49

….[By definition of logarithm]

iv. 102 = 0.01

– 2 = log10 (0.01) ….[By definition of logarithm]

2. Express the following in exponential form:

i. log9 6561 = 4 ii. log1/16 1

8

= 3

4

iii. log0.5 0.125 = 3 iv. log21

4

= 2.

Solution: i. log9 6561 = 4

94 = 6561 ….[By definition of logarithm]

ii. 1/16log1

8

= 3

4

3

41

16

= 1

8 ….[By definition of logarithm]

iii. log0.5 0.125 = 3

(0.5)3 = 0.125 ….[By definition of logarithm]

iv. log21

4

= 2

22 =1

4 ….[By definition of logarithm]

3. Find the values of: i. log1/2 8 ii. log5 0.008

iii. log5 3125 iv. log7 3 7 .

Solution: i. Let x = 1/2log 8

1

2

x

= 8 ….[By definition of logarithm]

(21)x = 23

2x = 23

x = 3

x = 3

1/2log 8 = 3

5


ii. Let x = log5 (0.008) 5x = 0.008 ….[By definition of logarithm]

5x = 8

1000

5x = 3

2

10

5x = 3

1

5

5x = 53 x = 3 log5 (0.008) = 3 iii. Let x = log5 3125 5x = 3125 ….[By definition of logarithm] 5x = 55 x = 5 log5 3125 = 5 iv. Let x = log7

3 7

7x = 3 7 ….[By definition of logarithm]

7x = 1

37

x = 1

3

log73 7 =

1

3 4. Simplify the following as single logarithm: i. log10 5 + 2 log10 4 ii. 2 log 7 log 14

iii. 1

2log5 36 + 2 log5 7

1

2log5 12

iv. log10 3 + log10 2 2log10 5

v. 2log 3 1

2log 16 + log 12.

Solution: i. log10 5 + 2 log10 4 = log10 5 + log10 4

2 ….[By exponent law] = log10 5 + log10 16 = log10 (5 16) ….[By product law] = log10 80 ii 2 log 7 log 14 = log 72 log 14 ….[By exponent law] = log 49 log 14

= log49

14

….[By quotient law]

= log7

2

iii. 1

2log5 36 + 2 log5 7

1

2log5 12

= 2 log57 + 1

2(log5 36 log512)

= 2 log5 7 + 1

2 log5

36

12


= 2 log5 7 + 1

2log5 3

= 25log 7 +

1

25log 3 ….[By exponent law]

= log5 49 + log5 3

= log5 49 3 ….[By product law] iv. log10 3 + log10 2 2 log10 5 = log10 (3 2) log10 5

2 ….[By product and exponent law]

= log10 6 log10 25

= log106

25


v. 2 log 3 1

2log 16 + log12

= log 32 log1

216 + log 12 ….[By exponent law]

= log 9 log 16 + log 12 = log 9 log 4 + log 12

= log9

4

+ log 12


= log9

124

….[By product law]

= log 27 5. Evaluate:

i. log5 4 25

625

ii. log10 12

5

+ log1025

21

log102

7

iii. log1015

16

+ log1064

81

log1020

27

iv. log102 + 16 log1016

15

+ 12 log1025

24

+ 7 log10

81

80

v. log10351

539

+2log1091

110

3log1039

110

.

6


Solution:

i. log5

4 25

625

= log5

1

425

625

= log5

1/4

2

4

5

5

= log5

14

25

= log5

7

25

= 7

2

log5 5 ….[By exponent law]

= 7

2

(1) …. alog a 1

= 7

2

ii. log1012

5

+ log1025

21

log102

7

= log1012 25

5 21

log10 2

7


= log10 20

7

log10 2

7

= log10

20

72

7


= 10log 10

= 1 …. alog a 1

iii. log1015

16

+ log1064

81

log1020

27

= log1015 64

16 81

log10 20

27


= log10 20

27

log1020

27

= log10

202720

27


= log10 1 = 0 …. alog 1 0

iv. log10 2 + 16 log1016

15

+ 12 log1025

24

+ 7 log1081

80

= log10 2 + log10

1616

15

+ log10

1225

24

+ log10

781

80

….[By exponent law]

= log10

16 12 7

16 12 7

16 25 812

15 24 80


= log10

16 12 74 2 4

16 12 7

2 5 32

(3 5) (3 8) (16 5)

= log10

16 12 74 2 4

16 16 12 12 7 7

2 5 32

3 5 3 8 16 5

= log10

64 24 28

16 16 12 712 3 4 7

2 5 32

3 5 3 2 2 5

= log10

64 24 28

16 16 12 36 28 7

2 5 32

3 5 3 2 2 5

= log10

64 24 28

28 64 23

2 5 32

3 2 5

= log10 (2 5) = log10 10

= 1 ….[ loga a = 1]

v. log10351

539

+ 2 log1091

110

3 log1039

110

= log10351

539

+ log10

291

110

log10

339

110


= log10 2

2

351 91

539 110

log10

339

110


= log10

2

2

3

3

351 91539 110

39

110


7


= log10

2 3

2 3

351 91 110

539 110 39

= log10

23

13 27 11013 7

11 49 13 3

= log10

32 2

2 3 3

13 3 11 1013 7

11 7 13 3

= log1010

= 1 .... alog a 1 6. Show that: i. log 360 = 3 log 2 + 2 log 3 + log 5

ii. log50

147

= log 2 + 2 log 5 log 3 2log 7

iii. log 75

16

2 log 5

9 + log

32

243

= log 2.

Solution: i. L.H.S. = log 360 = log (8 9 5) = log 8 + log 9 + log 5

….[By product law] = log 23 + log 32 + log 5 = 3 log 2 + 2 log 3 + log 5

….[By exponent law] = R.H.S.

ii. L.H.S. = log 50

147

= log 50 log 147 ….[By quotient law] = log (2 25) log (3 49) = log 2 + log 25 (log 3 + log 49)

….[By product law] = log 2 + log 52 (log 3 + log 72) = log 2 + 2 log 5 log 3 2 log 7

….[By exponent law] = R.H.S.

iii. L.H.S. = log 75

16

2 log5

9

+ log32

243

= log75

16

+ log32

243

log2

5

9


= log75 32

16 243

log25

81


= log5

4 5

25 3 2

2 3

log

25

81

= log4

25 2

3

log4

25

3

= log 4

4

25 2

325

3


= log 2 = R.H.S. 7. Solve for x: i. log (x + 3) + log (x 3) = log 16 ii. log (3x + 2) log (3x 2) = log 5

iii. 2 log10 x = 1 + log10 11

10

x

iv. log2 x + 1

2log2 (x + 2) = 2.

Solution: i. log (x + 3) + log (x – 3) = log 16 log [(x + 3)(x 3)] = log 16 ….[By product law] (x + 3) (x 3) = 16 x2 9 = 16 x2 = 25 x = 5 But log of negative number does not exist x 5 x = 5 ii. log (3x + 2) log (3x 2) = log 5

log 3 2

3 2

x

x= log 5 ….[By quotient law]

3 2

3 2

x

x= 5

3x + 2 = 15x 10 12 = 12x x = 1

iii. 2 log10 x = 1 + log1011

10x

log10 x2 log10

11

10x

= 1


log10 x2 log10

10 11

10

x

= 1

log10 2

10 11

10

xx

= 1 ….[By quotient law]

8


log10 210

10 11

x

x= 10log 10 …. alog a 1

210

10 11x

x= 10

2

10 11x

x= 1

x2 = 10x + 11 x2 10x 11 = 0 (x 11)(x + 1) = 0 x = 11 or x = 1

But log of negative number does not exist x 1 x = 11

iv. log2 x + 1

2log2 (x + 2) = 2

Multiplying throughout by 2, we get 2 log2 x + log2 (x + 2) = 4 log2 x

2 + log2 (x + 2) = 4 ….[By exponent law] log2 [x

2.(x + 2)] = 4 ….[By product law] x2(x + 2) = 24 ….[By definition of logarithm] x2 (x + 2) = 16 x3 + 2x2 16 = 0 (x 2)(x2 + 4x + 8) = 0 x 2 = 0 or x2 + 4x + 8 = 0

But x2 + 4x + 8 = 0 does not have real roots. x = 2

8. i. If log 3

x y

= 1

2 log x +

1

2 log y,

show that x y

y x = 7.

ii. If log 4

x y

= log x + log y ,

show that (x + y)2 = 20xy.

iii. If log 6

x y

= 1

2 (log x + log y),

show that x y

y x = 34.

Solution:

i. log 3

x y

=1

2log x +

1

2log y

Multiplying throughout by 2, we get

2 log 3

x y

= log x + log y

2 log3

x y= log xy ….[By product law]

log2

3

x y= log xy ….[By exponent law]

2

9

x y = xy

x2 + 2xy + y2 = 9xy x2 + y2 = 7xy Dividing throughout by xy, we get

x y

y x = 7

ii. log4

x y

= log x + log y

log4

x y

= log .x y ….[By product law]

log4

x y

= log xy

4

x y= xy

Squaring on both sides, we get

2

16

x y= xy

x2 2xy + y2 = 16xy Adding 4xy on both sides, we get x2 + 2xy + y2 = 20xy (x + y)2 = 20xy

iii. log6

x y

=1

2(log x + log y)


2 log6

x+ y= log x + log y

2 log 6

x y= log xy ….[By product law]

log2

6

x y= log xy ….[By exponent law]

2

36

x y= xy

x2 + 2xy + y2 = 36xy x2 + y2 = 34xy Dividing throughout by xy, we get

x y

y x = 34

9


9. i. If a2 + b2 = 3ab, show that

log a b

5

= 1

2 (log a + log b).

ii. If a2 + b2 = 7ab, prove that

2 loga b

3

= log a + log b.

iii. If a2 12ab + 4b2 = 0, prove that

log (a + 2b) = 1

2(loga + logb) + 2 log 2.

Solution: i. a2 + b2 = 3ab a2 + 2ab + b2 = 3ab + 2ab (a + b)2 = 5ab

2(a b)

5

= ab

2

a b

5

= ab

Taking log on both sides, we get

log 2

a b

5

= log ab

log2

a b

5

= log a + log b


2 loga b

5

= log a + log b


loga b

5

= 1

2(log a + log b)

ii. a2 + b2 = 7ab a2 + 2ab + b2 = 7ab + 2ab (a + b)2 = 9ab

2(a b)

9

= ab

2

a b

3

= ab


log2

a b

3

= log ab

log2

a b

3

= log a + log b


2 loga b

3

= log a + log b


iii. a2 12ab + 4b2 = 0 a2 + 4b2 = 12ab a2 + 4ab + 4b2 = 12ab + 4ab (a + 2b)2 = 16ab Taking log on both sides, we get log (a + 2b)2 = log (16ab) log(a + 2b)2 = log16 + log a + log b

….[By product law] log (a + 2b)2 = (log a + log b) + log 24 2 log (a + 2b) = (log a + log b) + 4 log 2

….[By exponent law] Dividing throughout by 2, we get

log (a + 2b) =1

2(log a + log b) + 2 log 2

10. Without using log table, prove the

following:

i. 1

4 < log10 2 <

1

3 ii.

2

5< log10 3 <

1

2

iii. 3

10< log10 2 <

1

3 iv.

2

3< log10 5 <

3

4

Solution:

i. We have to prove that, 1

4< log10 2 <

1

3

i.e., to prove that 1

4< log10 2 and log10 2 <

1

3

i.e., to prove that 1 < 4 log10 2 and 3 log10 2 < 1 i.e., to prove that 1 < log10 2

4 and log10 23 < 1

i.e., to prove that 410 10log 10 log 2

and 310 10log 2 log 10

.... alog a 1

i.e., to prove that 10 < 24 and 23 < 10 i.e., to prove that 10 < 16 and 8 < 10 which is true

1

4 < log10 2 <

1

3

ii. We have to prove that,2

5< log10 3 <

1

2


5< log10 3 and log10 3 <

1

2

i.e., to prove that 2 < 5 log10 3 and 2 log10 3< 1 i.e., to prove that 10 102log 10 < 5log 3

and 10 102log 3 < log 10

.... alog a 1

10


i.e., to prove that 2 510 10log 10 <log 3

and 210 10log 3 < log 10

i.e., to prove that 102 < 35 and 32 < 10


which is true

2

5< log10 3 <

1

2

iii. We have to prove that, 3

10< log10 2 <

1

3


10< log10 2 and log10 2 <

1

3

i.e., to prove that 3 < 10 log10 2 and 3 log10 2<1

i.e., to prove that 10103log 10 10log 2


.... alog a 1

i.e., to prove that 3 1010 10log 10 log 2




which is true

3

10< log10 2 <

1

3

iv. We have to prove that, 2

3< log10 5 <

3

4


3< log10 5 and log10 5 <

3

4

i.e., to prove that 2 < 3 log10 5 and 4 log10 5 < 3

i.e., to prove that 10 102log 10 3log 5

and 10 104log 5 3log 10

.... alog a 1

i.e., to prove that 2 310 10log 10 log 5

and 4 310 10log 5 log 10



which is true

2

3< log10 5 <

3

4

11. If log

b c

x

=

log

c a

y

=

log z

a b, show that

i. xyz = 1 ii. xaybzc = 1 iii. xb+c yc+a za+b = 1. Solution:

Let log

b cx

= log

c ay

=log z

a b= k

log x = k(bc), log y = k(ca), log z = k(ab) i. We have to prove that, xyz = 1 i.e., to prove that log(xyz) = log 1 i.e., to prove that log x + log y + log z = 0 L.H.S. = log x + log y + log z = k(b c) + k(c a) + k(a b) = k(b c + c a + a b) = 0 = R.H.S. ii. We have to prove that xa.yb.zc = 1 i.e., to prove that log (xa. yb. zc) = log 1 i.e., to prove that log xa + log yb + log zc = log 1 i.e., to prove that a log x + b log y + c log z = 0 L.H.S. = a log x + b log y + c log z = a.k(b c) + b.k(c a) + c.k(ab)

= k [a(b c) + b(c a) + c(a b)] = k (ab ac + bc ab + ac bc) = 0 = R.H.S. iii. We have to prove that xb + c. yc + a. za + b = 1 i.e., to prove that log (xb + c. yc + a. za + b) = log 1 i.e., to prove that log (x)b + c + log (y)c + a + log (z)a + b = 0 i.e., to prove that (b + c) log x +(c + a) log y + (a + b) log z = 0 L.H.S. = (b + c) log x + (c + a) log y + (a + b).log z = (b + c).k.(b c) + (c + a).k.(c a) + (a + b).k.(a b) = k (b2 c2 + c2 a2 + a2 b2) = 0 = R.H.S.

12. If log

a

x =

log

2

y =

log z

5= k and

x4y3z2 = 1, find ‘a’. Solution:

logLet

a

x=

log

2

y=

log z

5= k

log x = ak, log y = 2k, log z = 5k ….(i) But x4y3z2 = 1

11


Taking log on both sides, we get log (x4.y3.z2) = log 1 log x4 + log y3 + log z2 = 0 4 log x + 3 log y 2 log z = 0 4(ak) + 3(2k) 2(5k) = 0 ….[From(i)] 4ak + 6k 10k = 0 4ak = 4k a = 1

13. If a2 = b3 = c5 = d6, show that logd abc = 31

5.

Solution: a2 = b3 = c5 = d6 Taking log to the base d throughout, we get logd a

2 = logd b3 = logd c

5 = logd d6

2 logd a = 3 logd b = 5 logd c = 6 logd d 2 logd a = 3 logd b = 5 logd c = 6(1)

…. alog a 1

logd a =6

2= 3, logd b =

6

3= 2, logd c =

6

5

logd a + logd b + logd c = 3 + 2 +6

5

logd abc = 31

5

14. If ax = by = cz and b2 = ac, prove that

y = 2 z

z

x

x .

Solution: ax = by = cz Taking log to the base b throughout, we get logb a

x = logb by = logb c

z x logb a = y logb b = z logb c x logb a = y (1) = z logb c

logb a =y

x and logb c =

z

y ….(i)

Also, b2 = ac Taking log to the base b throughout, we get logb b

2 = logb (ac) 2 logb b = logb a + logb c

2 (1) =z

y y

x ….[From (i)]

2 = y1 1

z x

2 = yz

z

x

x

y = 2 z

zx

x

1.3 Change of Base law For any positive real numbers a, b and x

logb x = a

a

log

log b

x (x, b, a > 0, a, b ≠ 1)

Proof: Let logb x = m By definition of logarithm, we get x = bm Taking log on both sides to the base a, we get loga x = loga b

m loga x = mloga b

m = a

a

log

log b

x

Hence, logb x = a

a

log

log b

x

Corollary:

i. logb a = loga

log b ii. logb a =

a

1

log b

iii. logb a loga b = 1 Exercise 1.2 1. Evaluate the following:

i. 2

2

log 81

log 9

ii. 2

2

log 5

log 11 4

4

log 5

log 11

iii. 94

4 9

log 5log 7

log 5 log 7

iv. log25 5 log31

9

.

Solution:

i.

2

2

log 81

log 9 =

22

2

log 9

log 9

= 2

2

2log 9

log 9

= 2

ii. 2

2

log 5

log 11 4

4

log 5

log 11=

log 5

log 2log11

log 2

log 5

log 4log11

log 4

=log5

log11

log5

log11

= 0

12


iii. 4

4

log 7

log 5 9

9

log 5

log 7=

log 7 log 5

log 4 log9log5 log 7

log 4 log9

= log7 log5

log5 log7

= 1

iv. log25 5 log31

9

=

1log

log 5 9log 25 log3

=

1

2

2

log(5)

log(5)

1log(9 )

log3

=

1

2

2

log(5)

log(5)

2log(3)

log3

=

1log5

22log5

2log3

log3

= 1

24

= 1

2

2. Show that

i. logy x . logz y3. logx

3 2z = 1 ii. logb a

5 . logc b3 . loga c

7 = 105

iii. logv4 3u . logwv5. logu

5 4w = 3. Solution:

i. L.H.S. = logy x . logz y3. logx

3 2z

= logy

1

2x logz y3 logx

2

3z

= 1

2logy x 3 logz y

2

3logx z

= logy x logz y logx z

= log

log

x

y

log

log z

y

log z

log x = 1 = R.H.S. ii. L.H.S. = logb a

5. logc b3. loga c

7 = 5 logb a 3 logc b 7 loga c

= 5 3 7 loga

log b

log b

logc

logc

loga

= 105 = R.H.S.

iii. L.H.S. = logv4 3u . logwv5.logu

5 4w

= logv

3

4u logwv5 logu

4

5w

= 3

4logv u 5 logw v

4

5 logu w

= 3.log u

log v

log v

log w

log w

log u

= 3

= R.H.S. 3. If x = log5 7, y = log7 27, z = log3 5,

show that xyz = 3.

Solution:

x = log5 7, y = log7 27 and z = log3 5

L.H.S. = xyz = log5 7 . log7 27 . log3 5

= log7

log5

log 27

log7

log5

log3

=log 27

log3=

3log3

log3 =

3.log3

log3

= 3

= R.H.S.

4. Show that log6 7 = 2

2

log 7

1 log 3.

Solution:

L.H.S.= 6log 7

=

log7

log6

=

log7log 2log6log 2

= 2

2

log 7

log 6

= 2

2

log 7

log (2 3)

= 2

2 2

log 7

log 2 log 3

= 2

2

log 7

1 log 3 ….[ loga a = 1]

= R.H.S.

13


5. If log12 18 = x and log24 54 = y, prove that 5(x y) + xy = 1.

Solution: x = log1218

=log18

log12=

log 9 2

log 4 3

=

2

2

log 3 2

log 2 3

=2

2

log3 log 2

log 2 log3

x =2log3 log 2

2log 2 log3

y = log2454

=log54

log 24=

log 27 2

log 8 3

=

3

3

log 3 2

log 2 3

=3

3

log3 log 2

log 2 log3

y =3log3 log 2

3log 2 log3

L.H.S. = 5(x – y) + xy

L.H.S. = 52log3 log 2 3log3 log 2

2log 2 log3 3log 2 log3

+2log3 log 2 3log3 log 2

2log 2 log3 3log 2 log3

Let log 3 = a and log 2 = b

L.H.S. = 52a b 3a b

2b a 3b a

+2a b 3a b

2b a 3b a

= 5

2a b 3b a 2b a 3a b

2b a 3b a

+ 2a b 3a b

2b a 3b a

= 5 2 2 2 2

2 2

6ab 2a 3b ab 6ab 2b 3a ab

6b 2ab 3ab a

+2 2

2 2

6a 2ab 3ab b

6b 2ab 3ab a

= 52 2 2 2

2 2

6ab 2a 3b ab 6ab 2b 3a ab

6b 5ab a

+2 2

2 2

6a 5ab b

6b 5ab a

= 52 2

2 2

a b

6b 5ab a

+2 2

2 2

6a 5ab b

6b 5ab a

=2 2 2 2

2 2

5a 5b 6a 5ab b

6b 5ab a

=2 2

2 2

a 5ab 6b

a 5ab 6b

= 1 = R.H.S.

6. Prove that 5

1

log 3(3) = 5.

Solution:

L.H.S. = 5

1

log 3(3)

= 3log 53 …. ba

1log a

log b

= 5 …. aloga x x

= R.H.S.

7. Prove that 6 12 8

1 1 1

log 24 log 24 log 24 = 2.

Solution:

L.H.S. =6

1

log 24+

12

1

log 24+

8

1

log 24

= log24 6 + log2412 + log24 8

…. ba

1log a

log b

= log24 (6 12 8) = log24 576 = log24 (24)2 = 2 log24 24

= 2 1 ….[ loga a = 1] = 2 = R.H.S.

8. Prove that ab

1

log abc+

bc

1

log abc+

ca

1

log abc= 2.

Solution:

L.H.S. =ab

1

log abc+

bc

1

log abc+

ca

1

log abc

= logabc ab + logabc bc + logabc ca

…. ba

1log a

log b

= logabc (abbcca) = logabc (a

2b2c2)

14


= logabc (abc)2 = 2 logabc abc

= 2 1 ….[ loga a = 1] = 2 = R.H.S. 9. If x = 1 + loga bc, y = 1 + logb ca, z = 1 + logc ab,

then prove that xy + yz + zx = xyz. Solution: x = 1 + loga bc = loga a + loga bc x = loga abc y = 1 + logb ca = logb b + logb ca y = logb abc z = 1 + logc ab = logc c + logc ab z = logc abc

We have to prove that, xy + yz + zx = xyz

i.e., to prove that z z

1z

xy y x

xy

i.e., to prove that 1 1 11

z

x y

i.e., to prove that

1 1 11

z

x y

L.H.S. = 1 1 1

z

x y Substituting the values of x, y and z, we get

L.H.S. =a b c

1 1 1

log abc log abc log abc

= logabc a + logabc b + logabc c

…. ba

1log a

log b

= logabc abc = 1 ….[ loga a = 1] = R.H.S. 10. Prove that,

loga (x) + 22

alog x

+ 3a

3log x + …

+ pp

alog x = p loga x.

Solution:

L.H.S. = loga (x) + 22

alog x + 3

3loga

x + ….

+ pp

alog x

= log

log a

x+

2

2

log

loga

x+

3

3

log

loga

x+ … +

p

p

log

loga

x

= log

log a

x+

2.log

2.log a

x+

3.log

3.log a

x+…+

p log

p log a

x

= log

log a

x+

log

log a

x+

log

log a

x+ …. p times

= p.log

log a

x = p loga x

= R.H.S.

11. If log2 x + log4 x + log16 x = 21

4, find x.

Solution:

log2 x + log4 x + log16 x =21

4

log

log 2

x+

log

log 4

x+

log

log16

x=

21

4

2 4

log log log 21

log 2 4log(2) log(2)

x x x

log

log 2

x+

log

2.log 2

x+

log

4.log 2

x=

21

4

log

log 2

x 1 11

2 4

=21

4

log

log 2

x 7

4

=21

4

log

log 2

x= 3

log x = 3 log 2 log x = log 23

x = 23 x = 8 12. Solve for x, if x + log10 (1 + 2x) = x log10 5 + log10 6. Solution: x + log10 (1 + 2x) = x log10 5 + log10 6 x log10 10 + log10 (1+ 2x) = x log10 5 + log10 6 ….[ loga a = 1] log10 10x + log10 (1 + 2x) = log10 5

x + log10 6 log10 [10x.(1 + 2x)] = log10 (6 5x) 10x (1 + 2x) = 6 5x 2x 5x (1 + 2x) = 6 5x 2x (1 + 2x) = 6 Let 2x = a a.(1 + a) = 6 a + a2 = 6 a2 + a 6 = 0 (a + 3)(a 2) = 0 a + 3 = 0 or a 2 = 0 a = 3 or a = 2 Since 2x = 3, which is not possible 2x = 2 = 21 x = 1

15


1.4 Numerical Problems Common logarithms: Logarithm to the base 10 are called common logarithms. If log10 x = y, then x = 10y

Example: If we have to find log 42.36, we consider it as log10 (42.36) let log10 42.36 = x Then, 42.36 = 10x We know that it is not an easy job to find such x, and therefore we use the ready-made table to determine common logarithm of number. These tables are called tables of common logarithm. Characteristic and Mantissa of a Logarithm The common logarithm of any number consists of two parts, one of them is the integral part and the other is the fractional part. The integral part of a logarithm may be positive, negative or zero. This integral part is called the characteristic of logarithm. Thus the characteristic of logarithm may be positive, negative or zero. The fractional part of a logarithm is never negative. It is either positive or zero. This fractional part is called the mantissa of a logarithm. Thus the mantissa of a logarithm is never negative, it is either positive or zero. To determine the characteristic of log10 N: Characteristic of log10 N can be found by inspecting N. i. If N ≥ 1, then characteristic of log10 N is

always one less than the number of digits in the integral part of N.

Characteristic of log10 N

= (Number of digits in integral part of N) 1

For example,

Consider the number 14.3214, Since there are two digits in the integral part of 14.3214 its characteristic is 1.

Similarly characteristic of some more numbers are given below, where N ≥ 1

Number Characteristic

312.234 3 1 = 2

9.3214 1 1 = 0

12.3149 2 1 = 1

8704 4 1 = 3

870.4 3 1 = 2

12.34 2 1 = 1

1.234 1 1 = 0 ii. If N < 1, then the characteristic of log10 N is

always negative and one more than the number of zeros after the decimal point.

Characteristic of log10 N (N < 1)

= (number of zeros after decimal point + 1)

For example, consider the number 0.002341.

Since there are two zeros after the decimal

point, its characteristic is 3 .

Similarly, characteristics of some more numbers are given below, where N < 1

Number Characteristic

0.2341 (0 + 1) = 1 =1

0.01234 (1 + 1) = 2 = 2

0.008704 (2 + 1) = 3 = 3 Note: minus sign is written over the

characteristic and not before it. To determine the mantissa of log10 N: Mantissa is found using the log tables. Suppose if we want log10 128.3, then the characteristic is 2. For the mantissa, discard the decimal point and think of number 1283. Now we find the mantissa using the following table.

Mean Difference 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 8 12 17 21 25 29 33 37 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 8 11 15 19 23 26 30 34 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 3 7 10 14 17 21 24 28 31

16


In the above table, look for 12 in the first column and choose the corresponding row. Proceed along this row from left to right under the columns 0-9 till we reach the number under the column 8. This number is 1072. To this add the difference corresponding to 3 which is our fourth digit. This difference is 10. Thus, the required mantissa is 0.1082. Since the logarithm is characteristic + mantissa, log10 128.3 = 2 + 0.1082 = 2.1082 The mantissa of logarithm of numbers having the same sequence of digits are the same only, the characteristic change in such numbers. Number N Characteristic Mantissa log10N

314 2 0.4969 2.4969 3140 3 0.4969 3.4969

31400 4 0.4969 4.4969 31.4 1 0.4969 1.4969 3.14 0 0.4969 0.4969

0.314 1 0.4969 1.4969 0.0314 2 0.4969 2 .4969

0.00314 3 0.4969 3 .4969 To determine antilog10 M: In antilogarithm, we consider the problem of finding the number n when log n is known. If log n = m, we call n to be the antilog of m and write it as antilog m = n. Example:

log 100 = 2 antilog 2 = 100 To find the antilog of a number say M, it has to be expressed into two parts: i. characteristic i.e., integral part ii. mantissa i.e., decimal part Using the Antilogarithm table find the number corresponding to mantissa as done for finding logarithms from log tables. Next, we place the decimal point in this number such that its logarithm has the characteristic of M. The position of decimal point is decided from the characteristic. If the characteristic is p ≥ 0, then place the decimal point immediately after the (p + 1)th digit. Add zeros at the end if necessary.

If the characteristic is negative, say p, then we write (p 1) zeros in the beginning and place the decimal point before the zeros.

For example, consider the antilog of 2.4357. If we look up the mantissa of 0.4357 in the antilog table we get the digits 2727. Characteristic is 2. Hence the decimal point is to be put after (2 + 1) i.e., 3rd digit, which is 2 antilog 2.4357 = 272.7 Use of logarithmic table: i. Multiplication of two numbers: If x and y are two positive numbers, then by

law of logarithm, we have log (xy) = log x + log y xy = Antilog (log x + log y) Thus product of two numbers is antilogarithm

of sum of logarithms of these numbers. Example: 123.71 37.82 = Antilog (log 123.71 + log 37.82) = Antilog (2.0923 + 1.5777) = Antilog (3.67) = 4677 ii. Division of two numbers: If x and y are two positive numbers, then by

law of logarithm, we have

log

x

y = log x log y

x

y = Antilog (log x log y)

Example:

61.82

79222 = Antilog (log 61.82 log 79222)

= Antilog (1.7911 4.8988)

= Antilog ( 4 .8923) = 0.0007803 iii. Finding kth power of a positive number: If x is a positive number, then by law of

logarithm, we have log xk = k. log x xk = Antilog (k. log x) Example: (53.22)6 = Antilog [log (53.22)6] = Antilog [6 log (53. 22)] = Antilog (6 1.7261) = Antilog (10.3566) = 22730000000

17


iv. Finding nth root of given positive number: If x is a positive number and n is natural

number, then for finding nth root of x, we write

n x = x1/n , then x1/n = Antilog 1

logn

x

Example:

5 0.274 = (0.274)1/5 = Antilog [log (0.274)1/5]

= Antilog 1log 0.274

5

= Antilog 11.4378

5

= Antilog 15 4.4378

5

= Antilog [ 1 + 0.8876] = Antilog [ 1 .8876] = 0.7720 Exercise 1.3 1. If log10 3 = 0.4771212, without using log

tables, find

i. log10 9 ii. log10 3

iii. log10 1

9

iv. log10 (0.3).

Solution: i. log10 3 = 0.4771212 log10 9 = log10 (3

2) = 2.log103 = 2 (0.4771212) = 0.9542424 log10 9 = 0.9542

ii. log10 3 = log10

1

23

= 1

2(log103)

= 1

2(0.4771212)

= 0.2385606

log10 3 = 0.2386

iii. log101

9

= log10(9)1

= log1032

= 2 (log103) = 2 (0.4771212) = 0.9542424

log101

9

= 0.9542

iv. log10 (0.3) = log103

10

= log103 log1010

= 0.4771212 1

= 0.5228788

log10 (0.3) = 0.5229 2. Simplify:

i. 3 .5472 2 .8371 + 1.4581

ii. 1.2489 1 .0891 + 2 .8897.

Solution:

i. 3 .5472 2 .8371 + 1.4581

= 3.5472 1.4581 2 .8371

= 1.0053 2 .8371

= 0.1682

ii. 1.2489 1.0891 + 2 .8897

= (1.2489 + 2 .8897) 1.0891

= 0.1386 1.0891

= 1.0495 3. Find x, if

i. log10 x = 5 .2385 2.8031

ii. 4 log10 x = 3 1 .3345.

Solution:

i. log10 x = 5 .2385 2.8031

log10 x = 8 .4354

x = antilog (8 .4354) = 2.726 108

ii. 4 log10 x = 3 1.3345

4 log10 x = 3 ( 3 + 2.3345)

4 log10 x = ( 9 + 7.0035)

4 log10 x = 2 .0035

log10 x = 1

4( 2 .0035)

log10 x =1

4( 4 + 2.0035)

log10 x = (1+ 0.5009)

log10 x = 1.5009

x = antilog (1.5009) = 3.169 101 = 0.3169

18


4. Find the value of

2

1/3

0.3125

0.4629.

Solution:

Let x =

2

1/3

0.3125

0.4629


log10 x = log10

2

1/3

0.3125

0.4629

= log10(0.3125)2 log10 1/30.4629

= 2.log10 (0.3125) 1

3log10 (0.4629)

= 2.(1.4949) 1

3(1.6654)

= 2 ( 2 + 1.4949) 1

3( 3 + 2.6654)

= ( 4 + 2.9898) (1+ 0.8885)

= 2.9898 (1.8885 )

log10 x = 1.1013

x = antilog (1.1013) = 1.263 101 = 0.1263

5. Evaluate 22.41 2.61

1.374

.

Solution:

Let x = 22.41 2.61

1.374


log10 x = log10 22.41 2.61

1.374

= log10 (2.41)2 + log10 2.61 log10 1.374 = 2log10 2.41 + log10 2.61 log10 1.374 = 2.(0.3820) + 0.4166 0.1380 = 0.7640 + 0.4166 0.1380 log10 x = 1.1806 0.1380 = 1.0426 x = antilog (1.0426) = 1.104 101 = 11.04 6. Evaluate:

i. 316.23

426.8 ii.

35.87 0.0514

0.0578

.

Solution:

i. Let x = 316.23

426.8


log10 x = log10

1

316.23

426.8

=1

3log10

16.23

426.8

=1

3(log10 16.23 log10 426.8)

= 1

3(1.2103 2.6302)

= 1

3( 2 .5801)

= 1

3( 3 + 1.5801) = (1+ 0.5267)

log10 x = 1.5267

x = antilog (1.5267) = 3.362 101 = 0.3362

ii. Let x =35.87 0.0514

0.0578


log10 x = log10

1

235.87 0.0514

0.0578

= 1

2log10

35.87 0.0514

0.0578

= 1

2(log10 35.87 + log10 0.0514

log10 0.0578)

= 1

2(1.5548 + 2 .7110 2 .7619)

= 1

2(0.2658 2 .7619)

= 1

2(1.5039)

log10 x = 0.7519 x = antilog (0.7519) = 5.648

7. Find the value of 3

5

27.38 0.3052

31.65 0.3028

.

Solution:

Let x =3

5

27.38 0.3052

31.65 0.3028


log10 x = log10

13

15

27.38 (0.3052)

31.65 0.3028

19


= log10 27.38 +1

3log10 0.3052

(log10 31.65 +1

5log10 0.3028)

= 1.4375 +1

3(1.4846)

11.5004 (1.4811)

5

= 1.4375 +1

3( 3 + 2.4846)

1

(1.5004) (5 4.4811)5

= 1.4375 + (1+ 0.8282) [(1.5004)

+ (1+ 0.8962)]

= 1.4375 + 1.8282 (1.5004 + 1.8962)

= 1.2657 1.3966

logx 10 = 1.8691

x = antilog (1.8691) = 7.398 101 x = 0.7398 8. Find the value of x, if

x =

5

43

72.14 45

2.8 32

.

Solution:

x =

5

43

72.14 45

2.8 32


log10 x = log10

11 4

5 2

13 2

(72.14) (45)

(2.8) (32)

= 1

4log10

15 2

13 2

(72.14) (45)

(2.8) (32)

= 1

4[5 log10 72.14 +

1

2log10 45

(3log10 2.8 +1

2log10 32)]

= 1

41

5(1.8581) (1.6532)2

1

3 0.4472 1.50512

= 1

4[9.2905+ 0.8266 (1.3416 + 0.7526)]

= 1

4(10.1171 2.0942)

= 1

4(8.0229)

log10 x = 2.0057

x = antilog (2.0057) = 101.4 9. Given a = 37.58, b = 25.56 and c = 42.29,

find the value of p, if p = 3

5 2

a

c b.

Solution:

p = 3

5 2

a

c .b=

3

5 2

(37.58)

(42.29) (25.56)=

3

2

5 2

37.58

42.29 25.56


log10 p = log10

3

2

5 2

37.58

42.29 25.56

= log10 3

2 5 210 1037.58 log 42.29 log 25.56

= 3

2 log10 (37.58) (5 log10 42.29 + 2 log10 25.56)

= 3

2(1.5749) (5 1.6262 + 2 1.4075)

= 4.7247

2 (8.1310 + 2.8150)

= 2.3624 10.9460

log10 p = 9 .4164

p = antilog ( 9 .4164) = 2.608 109

10. Given = 3.142, r = 2.307, h = 8.5. Find the

value of V, if V = r2h. Solution:

V = r2.h

V = 3.142 (2.307)2 8.5 Taking log on both sides, we get log10V = log10 3.142 + log10 (2.307)2 + log10 8.5 = log10 3.142 + 2log10 (2.307) + log10 8.5 = 0.4972 + 2 (0.3630) + 0.9294 = 0.4972 + 0.7260 + 0.9294

log10V= 2.1526

V = antilog (2.1526) = 1.421 102 = 142.1

20


11. Find the value of 2 33 35.285 23.45 .

Solution:

Let x = 2 33 35.285 23.45

Let a = (35.285)2 Taking log on both sides, we get log10 a = 2 log10 (35.285) = 2 (1.5476) log10 a = 3.0952 a = antilog (3.0952) = 1.246 103 = 1246 Let b = (23.45)3 Taking log on both sides, we get log10 b = 3 log10 (23.45) = 3(1.3701) = 4.1103 b = antilog (4.1103) = 1.289 104 = 12890

x = 3 a b = 3 1246 12890 = 3 14136 Taking log on both sides, we get

log10x = 1

3log10 (14136) =

1

3 (4.1501)

log10 x = 1.3834 x = antilog (1.3834) = 24.17 12. If log 33.48 = 1.5247854, find log 3 33.48 ,

log 334800, antilog 4.5247854 without using log tables.

Solution: Given, log 33.48 = 1.5247854

log 3 33.48 = log 1

333.48

= 1

3 log 33.48 =

1

3 (1.5247854)

log 3 33.48 = 0.5082618 log 334800 = 5.5247854 antilog (4.5247854) = 3.348 104 = 33480 Miscellaneous Exercise – 1 1. For any base show that

log (1 + 2 + 3) = log 1 + log 2 + log 3. Solution: L.H.S. = log (1 + 2 + 3) = log 6 = log (1 2 3) = log 1 + log 2 + log 3 = R.H.S. 2. Find x, if x = 33log 23 Solution:

x = 33log 23 = 33log 23

x = (2)3 ….. aloga x = x

x = 8

3. Show that,

log 2 1x x + log 2 1x x

= 0.

Solution:

L.H.S. = log 2 1 x x + log 2 1

x x

= log 2 21 . 1 x x x x

= log 22 21

x x

= log (x2 + 1 x2)

= log 1 = 0 ....[ loga1 = 0] = R.H.S.

4. Show that, log 2 2 2a b c

log logbc ca ab

= 0.

Solution:

L.H.S. = log 2a

bc + log

2b

ca+ log

2c

ab

= log 2 2 2a b c

bc ca ab

= log 2 2 2

2 2 2

a b c

a b c

= log 1

= 0 ....[ loga1 = 0] = R.H.S. 5. Simplify, log (log x4) log (log x). Solution: log (log x4) log (log x) = log (4.log x) log (log x) = log4 + log (log x) log (log x) = log4 6. Simplify,

log1028

45

log1035

324

+ log10325

432

log1013

15

Solution:

log1028

45

log10 35

324

+ log10325

432

log1013

15

= 28 325

log log10 1045 432

35 13

log log10 10324 15

21


= log1028 325

45 432

log10 35 13

324 15

= log107 65

9 108

log10 7 13

3 324

= log10

7 659 1087 13

3 324

= log107 65 3 324

9 108 7 13

= log10 5

7. If loga b 1

2 2

(log a + log b),

then show that a = b. Solution:

loga b

2

= 1

2 (log a + log b)

2 log a b

2

= log a + log b

log 2

a b

2

= log ab

2a b

4

= ab

a2 + 2ab + b2 = 4ab a2 + 2ab 4ab +b2 = 0 a2 2ab + b2 = 0 (a b)2 = 0 a b = 0 a = b 8. If b2 = ac, prove that log a + log c = 2 log b. Solution: b2 = ac Taking log on both sides, we get log b2 = log ac 2 log b = log a + log c log a + log c = 2 log b 9. Solve for x, logx (8x 3) logx 4 = 2. Solution: logx (8x 3) logx 4 = 2

logx 8 3

4

x

= 2

x2 = 8 3

4

x

4x2 = 8x 3

4x2 8x + 3 = 0 4x2 2x 6x + 3 = 0 2x(2x 1) 3 (2x 1) = 0 (2x 1) (2x 3) = 0 2x 1 = 0 or 2x 3 = 0

x = 1

2 or x =

3

2 10. If a2 + b2 = 7ab, show that

log a b

3

= 1

2 log a +

1

2 log b.

Solution: a2 + b2 = 7ab a2 + 2ab + b2 = 7ab + 2ab (a + b)2 = 9ab

2(a b)

9

= ab

2

a b

3

= ab


log2

a b

3

= log (ab)

2 loga b

3

= log a + log b

Dividing throughout by 2, we get

loga b 1

3 2

log a + 1

2log b

11. If log1

5 2

x y

log x + 1

2 log y,

show that x2 + y2 = 27xy. Solution:

log5

x y

= 1

2 (log x) +

1

2(log y)


2 log 5

x y= log x + log y

log2

5

x y = log xy

2

25

x y = xy

x2 2xy + y2 = 25 xy x2 + y2 = 27xy

22


12. If log3 [log2 (log3 x)] = 1, show that x = 6561. Solution: log3 [log2(log3 x)] = 1 log2 (log3 x) = 31 log3 x = 23 log3 x = 8 x = 38 x = 6561 13. Without using log tables, prove that

2

5 < log10 3 <

1

2.

Solution: Refer solution Ex. 1.1 Q.10 (ii) 14. Show that

7log 15

16

+ 6log 8

3

+5log 2

5

+log 32

25

= log 3.

Solution:

L.H.S. = 7 log15

16

+6log8

3

+ 5log2

5

+ log32

25

= log7

15

16

+ log6

8

3

+ log5

2

5

+ log 32

25

= log7

4

3 5

2

+ log632

3

+ log5

2

5

+ log5

2

2

5

= log7 7

28

3 5

2

+ log18

6

2

3

+ log5

5

2

5

+ log5

2

2

5

= log7 7 18 5 5

28 6 5 2

3 5 2 2 2

2 3 5 5

= log 3 = R.H.S.

15. Solve: 42log x + 4 log4

2

x= 2.

Solution:

42log x + 4 log4

2

x = 2

42log x + 4 log4

1

22

x

= 2

24log x +4

2log4

2

x

= 2

2 2log x + 2 log42

x

= 2

2log x + log42

x

= 1

2log x +

2

2

2log

log 4

x

= 1

2log x +

2

22

2log

log (2)

x

= 1

2log x + 2 2

2

log 2 log

2log 2

x= 1

2log x + 21 log

2(1)

x= 1 ….[ loga a = 1]

Let log2 x = a

1 a

a2

= 1


2 a + 1 a = 2

2 a = a + 1 Squaring on both sides, we get 4a = (a + 1)2 4a = a2 + 2a + 1 a2 2a + 1 = 0 (a 1)2 = 0 a 1 = 0 a = 1 Since log2 x = a log2 x = 1 x = 21 x = 2 16. Find the value of

10

10 10

3 log 343

1 49 1 12 log log

2 4 2 25

.

Solution:

10

10 10

3 log 3431 49 1 1

2 log log2 4 2 25

= 3

101 1

2 210 10

3 log 7

49 12 log log

4 25

= 10

10 10

3 3.log 77 1

2 log log2 5

23


= 10

10

3 1 log 7

7 12 log

2 5

= 10

10

3 1 log 7

72 log

10

= 10

10 10

3 1 log 7

2 log 7 log 10

= 10

10

3 1 log 7

2 log 7 1

….[loga a =1]

= 10

10

3 1 log 7

1 log 7

= 3

17. If loga logb logc

2z z 2 z 2x y y x x y

,

show that abc = 1. Solution:

Let log a

2z x y=

log b

z 2 y x=

log c

z 2 x y = k

log a = k (x + y 2z), log b = k (y + z 2x), log c = k (z + x 2y) We have to prove that abc = 1 i.e., to prove that log (abc) = log 1 i.e. , to prove that log a + log b + log c = 0 L.H.S.= log a + log b + log c = k (x + y 2z) + k (y + z 2x) + k (z + x 2y) = k (x + y 2z + y + z 2x + z + x 2y) = 0 = R.H.S. 18. Show that logy x

3 . logz y4 . logx z

5 = 60. Solution: L.H.S. = logy x

3. logz y4. logx z

5 = 3 logy x 4 logz y 5 logx z

= 60 log

log

x

y

log

log z

y

log z

log x

= 60(1) = 60 = R.H.S.

19. If 2 2 2log a log b log c

4 6 3k and a3b2c = 1,

find the value of k. Solution:

Let 2log a

4= 2log b

6= 2log c

3k = x

log2 a = 4x, log2 b = 6x, log2 c = 3k.x ….(i) Also, a3b2c = 1

Taking log to the base 2 throughout, we get log2 (a

3b2c) = log2 1 log2 a

3 + log2 b2 + log2 c = 0

3 log2 a + 2 log2 b + log2 c = 0 3(4x) + 2(6x) + 3kx = 0 ….[From (i)] 12x + 12x + 3kx = 0 3kx = 24x k = 8

20. If a2 = b3 = c4 = d5, show that loga (bcd) =47

30.

Solution: a2 = b3 = c4 = d5 Taking log to the base a throughout, we get loga a

2 = loga b3 = loga c

4 = loga d5

2 loga a = 3 loga b = 4 loga c = 5 loga d 2(1) = 3 loga b = 4 loga c = 5 loga d

loga b = 2

3, loga c =

2 1

4 2 and loga d =

2

5

loga b + loga c + loga d = 2 1 2

3 2 5

loga bcd = 47

30

21. Using log tables, evaluate

5 3

4 2

(2.3) (0.537)

(72.5) (18.25)

.

Solution:

Let x = 5 3

4 2

(2.3) + (0.537)

(72.5) (18.25)

Let a = (2.3)5 Taking log on both sides, we get log10 a = 5.log10 (2.3) log10 a = 5 (0.3617) = 1.8085 a = antilog (1.8085) = 6.434 101 = 64.34 b = (0.537)3 Taking log on both sides, we get

log10 b = 3.log10 (0.537) = 3(1.7300)

= 3( 3 + 2.7300) = ( 9 + 8.1900) = 1.1900

b = antilog (1.1900) = 1.549 101 = 0.1549 c = (72.5)4 Taking log on both sides, we get log10 c = 4.log10 (72.5) = 4(1.8603) = 7.4412 c = antilog (7.4412) = 2.762 107

d = (18.25)2

24


Taking log on both sides, we get log10 d = 2.log10 (18.25) = 2(1.2613) = 2.5226 d = antilog (2.5226) = 3.332 102 = 333.2

x = a b

c d

= 64.34 0.1549

27620000 333.2

= 64.4949

27619666.8


log10 x = log10

1

264.4949

27619666.8

= 1

2(log10 64.4949 log10 27619666.8)

= 1

2(1.8095 7.4411)

= 0.9048 3.7206

log10 x = 3 .1842

x = antilog ( 3 .1842) = 1.529 103

= 0.001529 22. If log10 2 = 0.3010, find the number of digits

in 264. Solution: log10 2 = 0.3010 log10 2

64 = 64(log10 2) = 64(0.3010) = 19.264 Characteristic = 19 No. of digits in 264 is (19 + 1) = 20. 23. If area of a circle is 88.2 sq.m and = 3.142,

find r, correct to two significant figures. Solution: Given, area of a circle = 88.2 sq.m, = 3.142 But, area of a circle = r2 88.2 = 3.142 r2 Taking log on both sides, we get log10 88.2 = log10 3.142 + log10 r

2 = log10 3.142 + 2 log10 r 2 log10 r = log10 88.2 log10 3.142 = 1.9455 0.4972 = 1.4483 log10 r = 0.72415 = 0.7242 r = antilog (0.7242) = 5.299 = 5.30 m 24. The population of a town at present is

80000. If the annual rate of increase is 4%, find the population after 4 years.

(Use the formula: A = Pn

r1

100

, where

P is the original population.)

Solution: Given, P = 80000, r = 4%, n = 4 years

A = Pn

r1

100

A = 800004

41

100

= 800004

104

100

Taking log on both sides, we get log10 A = log10 80000 + 4 (log10 104 log10 100) log10 A = 4.9031 + 4(2.0170 2) = 4.9031 + 4(0.0170) = 4.9031 + 0.0680 = 4.9711 A = antilog (4.9711) = 9.356 104

= 93560

Additional Problems for Practice Based on Exercise 1.1 1. Write the following in the logarithmic form: i. 25 = 32

ii. 3

216 = 64

iii. 53 = 1

125

iv. 23 = 0.125 2. Express the following in the exponential form: i. log3 2187 = 7

ii. log1/27 1

9

= 2

3

iii. log10 (0.001) = 3

iv. log6 1

36

= 2

3. Find the values of: i. log1/9 729 ii. log8 16

iii. log0.5 1

16

iv. log5 5 5 4. Simplify the following as single logarithm: i. log103 + 2log10 5 ii. 2 log 9 log 15

25


iii. log 8 + 2 log 4 1

2log 64

iv. log1016

15

+ log1025

24

+ 2log10(3)

v. log729

64

+ 2log8

3

2log 9

5. Show that: log 450 = log 2 + 2log 3 + 2log 5. 6. Solve for x: i. log(x + 2) + log(x 2) = log 5 ii. 2log2 x log2(x + 3) = 2

7. i. If log(x y) = 1

2(log x + log y), show

that x2 + y2 = 3xy.

ii. If log3

x y =

1

2(log x + log y), show

that x2 + y2 = 11xy. 8. i. If x2 + y2 = 23xy, show that

log5

x y =

1

2(log x + log y)

ii. If a2 + b2 = 4ab, show that

log a b

6

= 1

2(log a + log b)

9. Without using log table, show that:

i. 1

3 < log10 3 <

1

2

ii. 3

5 < log104 <

2

3

10. If log log log z

a 4 5

x y= k and x3y2z7 = 1,

find a.

11. If a2 = b4 = c7 = d8, show that logd abc = 64

7.

Based on Exercise 1.2 1. Evaluate the following:

i. 2

2

log 64

log 4

ii. 3 4

3 4

log 5 log 5

log 7 log 7

iii. log42 log23 log35 log54

iv. log164 log41

32

2. Show that:

i. logy 4 x logz y

5 logx 5 4z = 1

ii. logy x3 logz y

2 logx z4 = 24

iii. log34 log5 27 log4 5 = 3 3. If x = log164, y = log432, z = log642, show that

xyz = 5

24.

4. Show that log105 = 2

2

log 5

1 log 5.

5. Prove that:

i. 1

log 575 = 7

ii. 2 3 7 42

1 1 1 1

log log log log

x x x x

iii. 1

log zx xy+

1

log zy xy+

z

1

log zxy = 1

6. If log8 x + log4 x + log2 x = 11, find x. 7. Prove that: log2 3 + log4 9 + log8 27 = 3log2 3. Based on Exercise 1.3 1. If log10 7 = 0.845098, without using log table,

find

i. log10 49 ii. log10 7

iii. log101

49

iv. log10 (0.7)

2. Simplify:

i. 3.5275 2.4376 1.4076

ii. 1.4375 2.7398 1.3701 3. Find x, if 5log10 x = 4 1.3861 .

4. Evaluate: 53.22 0.5472

0.02489

5. Evaluate the following:

i. 3

24.395 3.16

8.79

ii. 5.38 0.47

6. Find the value of x, if

x = 3

5

28.45 0.3254

32.43 0.3046

.

7. If log 28.45 = 1.454082, find log 3 28.45 ,

log 28450, antilog 2.454082 , without using log tables.

26


Based on Miscellaneous Exercise 1 1. Find x, if x = 3log 433 . 2. Show that:

4 4 4

2 2 2 2 2 2

a b clog log log

b c c a a b

= 0.

3. Show that:

3 log 81

80 + 5 log

25

24 + 7 log

16

15 = log 2.

4. If x = e e

e e

y y

y y

, show that y = e

1 1log

2 1

x

x

.

5. Solve for x : log 2 + log (x + 3) log (3x 5) = log 3. 6. If x2 + y2 = 47xy, show that

log7

x y =

1

2(log x + log y).

7. If log6

x y =

1

2(log x + log y), show that

x2 + y2 = 38xy. 8. If log2[log3(log2x)] = 1, show that x = 512. 9. Without using log table, prove that

3

5< log105 <

3

4.

10. Prove that:

z

1 1 1

1 log z 1 log z 1 log

x yy x xy= 1.

11. Show that : logb a

4.logc b7.loga c

9 = 252. 12. Find the value of

log5 1

15

+ log5 1

16

+ log5 1

17

+….. + log5 1

1624

13. If a = log24 12, b = log36 24 and c = log48 36,

prove that abc = 2bc 1. 14. If loga b + logc b = 2(loga b) (logc b), prove that

b2 = ac. 15. If ax = by = cz = dw(d 1), then show that

logaabcd = 1 1 1

z w

x

x y.

16. If log103 = 0.4771212, find the number of

digits in 321.

Multiple Choice Questions

1. For y = loga x to be defined 'a' must be (A) Any positive real number (B) Any number (C) e (D) Any positive real number 1

2. log91

243

=

(A) 5

2

(B) –5

(C) 2

5 (D)

5

2

3. The number log4 5 is (A) An integer (B) A rational number (C) An irrational number (D) A prime number 4. Logarithms to the base 10 are called (A) common logarithms (B) complex logarithms (C) natural logarithms (D) artificial logarithms 5. If log45 = x and log5 6 = y, then log2 3 = (A) 2xy + 1 (B) 2xy 1 (C) 2x + 1 (D) 2y + 1 6. The value of log3 4log4 5log5 6log6 7log7 8log8 9

is (A) 1 (B) 2 (C) 3 (D) 4

7. 7 log16

15

+ 5log25

24

+ 3 log81

80

is equal

to (A) 0 (B) 1 (C) log 2 (D) log 3 8. If log5 a.loga x = 2 then x is equal to (A) 125 (B) a2 (C) 25 (D) None of these 9. If A = log2 log2 log4 256 + 2

2log 2, then A is

equal to (A) 2 (B) 3 (C) 5 (D) 7 10. If a = log24 12,b = log36 24 and c = log48 36,

then 1 + abc is equal to (A) 2ab (B) 2ac (C) 2bc (D) 0

27


11. If log10 2 = 0.30103,log10 3 = 0.47712, then the number of digits in 312 + 28 is

(A) 7 (B) 8 (C) 9 (D) 10 12. The solution of the equation

log7 log5 2( 5 )x x = 0 is

(A) x = 2 (B) x = 3 (C) x = 4 (D) x = 2

13. If 10log a

2= 10log b

3= 10log c

5, then bc =

(A) a (B) a2 (C) a3 (D) a4

14. The value of1

log 1024252 is

(A) 1

4 (B) –4

(C) 4 (D) 1

4

15. If x = logb a, y = logc b, z = loga c, , then xyz is (A) 0 (B) 1 (C) 3 (D) None of these

16. 3

1

log x+

4

1

log x+

7

1

log x=

(A) log14 x (B) log84 x

(C) 84

1

log x (D)

14

1

log x

17. If logx : logy : logz = (y z):(z x):(x y),

then (A) xy.yz.zx= 1 (B) xxyyzz = 1

(C) z zx yx y = 1 (D) None of these

18. b b

1

1 log a log c +

c c

1

1 log a log b +

a a

1

1 log b log c

is equal to

(A) abc (B) 1

abc

(C) 0 (D) 1 19. If logx (27)1 = 3, then the value of x = (A) 3 (B) 3 (C) 9 (D) 9 20. loga 21 loga 3 loga 7 = (A) 0 (B) 1 (C) 2 (D) 1

21. If log (x y) = 1

2(log x + log y), then x2 + y2 =

(A) xy (B) 2xy (C) 0 (D) 3xy

22. Evaluate29.5 67.8 39.3

57.55

(A) 217.8 (B) 220.8 (C) 215.6 (D) 10.87 23. Cube root of 0.3813 is (A) 0.3858 (B) 725.1 (C) 38.58 (D) 0.7251

24. 1.287

24.56 36.62 =

(A) .4291 (B) 1.4291

(C) .04291 (D) 2 .4291

25. 2/3 3/214 4.265

17.8

=

(A) 1212 (B) 12.12 (C) 11.12 (D) 10.12

Answers to Additional Practice Problems Based on Exercise 1.1 1. i. 5 = log2 32

ii. 3

2 = log16 64

iii. 3 = log5 1

125

iv. 3 = log2 (0.125) 2. i. 37 = 2187

ii.

2

31 1

27 9

iii. 103 = 0.001

iv. 62 = 1

36

3. i. 3

ii. 4

3

iii. 4

iv. 3

2

28


4. i. log1075

ii. log27

5

iii. log 16 iv. 1 v. 0 6. i. 3 ii. 6 10. 9 Based on Exercise 1.2 1. i. 3 ii. 0

iii. 1 iv. 5

4

6. 64 Based on Exercise 1.3 1. i. 1.690196 ii. 0.422549 iii. 1.690196 iv. 0.154902 2. i. 0.4975 ii. 0.8072 3. 0.3228 4. 1170 5. i. 87.60 ii. 1.591 6. 0.7656 7. 0.484694, 4.454082, 0.028450 Based on Miscellaneous Exercise – 1 1. 64 5. 3 12. 3 16. 11

Answers to Multiple Choice Questions

1. (D) 2. (A) 3. (C) 4. (A)

5. (B) 6. (B) 7. (C) 8. (C)

9. (C) 10. (C) 11. (C) 12. (C)

13. (D) 14. (C) 15. (B) 16. (C)

17. (B) 18. (D) 19. (B) 20. (A)

21. (D) 22. (A) 23. (D) 24. (C)

25. (B)

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