std 11 commerce, mathematics and statistics - ii ... · • topic-wise distribution of all textual...
TRANSCRIPT
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
© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical
including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.
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.
Please write to us on: [email protected]
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
(Based on Miscellaneous) Q.1
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
(Based on Miscellaneous) Q.9
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
(Based on Miscellaneous) Q.16
3
Chapter 01: Logarithms
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
Std. XI : Commerce (Maths ‐ II)
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
Chapter 01: Logarithms
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
….[By quotient law]
= 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
….[By quotient law]
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
….[By quotient law]
= 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
Std. XI : Commerce (Maths ‐ II)
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
….[By product law]
= log10 20
7
log10 2
7
= log10
20
72
7
….[By quotient law]
= 10log 10
= 1 …. alog a 1
iii. log1015
16
+ log1064
81
log1020
27
= log1015 64
16 81
log10 20
27
….[By product law]
= log10 20
27
log1020
27
= log10
202720
27
….[By quotient law]
= 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
….[By product law]
= 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
….[By exponent law]
= log10 2
2
351 91
539 110
log10
339
110
….[By product law]
= log10
2
2
3
3
351 91539 110
39
110
….[By quotient law]
7
Chapter 01: Logarithms
= 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
….[By exponent law]
= log75 32
16 243
log25
81
….[By product law]
= 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
….[By quotient law]
= 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
….[By exponent law]
log10 x2 log10
10 11
10
x
= 1
log10 2
10 11
10
xx
= 1 ….[By quotient law]
8
Std. XI : Commerce (Maths ‐ II)
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)
Multiplying throughout by 2, we get
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
Chapter 01: Logarithms
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
….[By product law]
2 loga b
5
= log a + log b
….[By exponent law]
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
Taking log on both sides, we get
log2
a b
3
= log ab
log2
a b
3
= log a + log b
….[By product law]
2 loga b
3
= log a + log b
….[By exponent law]
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
i.e., to prove that 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
Std. XI : Commerce (Maths ‐ II)
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
i.e., to prove that 100 < 243 and 9 < 10
which is true
2
5< log10 3 <
1
2
iii. We have to prove that, 3
10< log10 2 <
1
3
i.e., to prove that 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
and 10 103log 2 log 10
.... alog a 1
i.e., to prove that 3 1010 10log 10 log 2
and 310 10log 2 log 10
i.e., to prove that 103 < 210 and 23 < 101
i.e., to prove that 1000 < 1024 and 8 < 10
which is true
3
10< log10 2 <
1
3
iv. We have to prove that, 2
3< log10 5 <
3
4
i.e., to prove that 2
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
i.e., to prove that 102 < 53 and 54 < 103
i.e., to prove that 100 < 125 and 625 < 1000
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
Chapter 01: Logarithms
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
Std. XI : Commerce (Maths ‐ II)
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
Chapter 01: Logarithms
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
Std. XI : Commerce (Maths ‐ II)
= 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
Chapter 01: Logarithms
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
Std. XI : Commerce (Maths ‐ II)
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
Chapter 01: Logarithms
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
Std. XI : Commerce (Maths ‐ II)
4. Find the value of
2
1/3
0.3125
0.4629.
Solution:
Let x =
2
1/3
0.3125
0.4629
Taking log on both sides, we get
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
Taking log on both sides, we get
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
Taking log on both sides, we get
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
Taking log on both sides, we get
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
Taking log on both sides, we get
log10 x = log10
13
15
27.38 (0.3052)
31.65 0.3028
19
Chapter 01: Logarithms
= 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
Taking log on both sides, we get
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
Taking log on both sides, we get
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
Std. XI : Commerce (Maths ‐ II)
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
Chapter 01: Logarithms
= 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
Taking log on both sides, we get
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)
Multiplying throughout by 2, we get
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
Std. XI : Commerce (Maths ‐ II)
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
Multiplying throughout by 2, we get
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
Chapter 01: Logarithms
= 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
Std. XI : Commerce (Maths ‐ II)
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
Taking log on both sides, we get
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
Chapter 01: Logarithms
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
Std. XI : Commerce (Maths ‐ II)
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
Chapter 01: Logarithms
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
Std. XI : Commerce (Maths ‐ II)
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)