KSEEB Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1

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Karnataka State Syllabus Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1

Question 1.
Find the value of :
(i) 26
(ii) 93
(iii) 112
(iv) 54
Solution:
i) 26
26 = 2 × 2 × 2 × 2 × 2 × 2 = 64

ii) 93
93 = 9 × 9 × 9 = 729

iii) 112
112 = 11 × 11 = 121

iv) 54
54 = 5 × 5 × 5 × 5 = 625.

Question 2.
Express the following in exponential form :
(i) 6 × 6 × 6 × 6
(ii) t × t
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c × c × d
Solution:
i) 6 × 6 × 6 × 6 = 64

ii) t × t = t2

iii) b × b × b × b = b4

iv) 5 × 5 × 7 × 7 × 7 = 52 × 73

v) 2 × 2 × a × a = 22 × a2

vi) a × a × a × c × c × c × c × d = a3 × c4 × d

KSEEB Solutions

Question 3.
Express each of the following numbers using exponential notation :
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Solution:
i) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
The exponential notation of 512 is 29

ii) 343 = 7 × 7 × 7 = 73
∴ The exponential notation of 343 is 73

iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
∴ The exponential notation of 729 is 36

iv) 3125 = 5 × 5 × 5 × 5 × 5 = 55
∴ The exponential notation of 3125 is 55

Question 4.
Identify the grater number, wherever possible, in each of the following ?
Solution:
i) 43 or 34
43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
∴ 34 > 43

ii) 53 or 35
53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
∴ 35 >53

iii) 28 or 82
28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
∴ 28 > 82

iv) 1002 or 2100
1002 = 100 × 100= 10,000
28 = (210)10 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
× 2 × 2
= (1024)10 = [(1024)2]5 = [10,48,576]
= (10,48,476)5 > 10,000
= 2100 > 1002

KSEEB Solutions

v) 210 or 102
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
102 = 10 × 10 = 100
∴ 210 > 102

Question 5.
Express each of the following as product of powers of their prime factors :
Solution:
i) 648
648 = 2 × 2 × 2 × 3 × 3 × 3 × 3
= 23 × 34

ii) 405
405 = 5 × 3 × 3 × 3 × 3
= 5 × 34

iii) 540
540 = 2 × 2 × 3 × 3 × 3 × 5
= 22 × 33 × 5

iv) 3,600
3,600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 24 × 32 × 52

Question 6.
Simplify :
(i) 2 × 103
(ii) 72 × 22
(iii) 23 × 5
(iv) 3 × 44
(v) 0 × 102
(vi) 52 × 33
(vii) 24 × 32
(viii) 32 × 104
Solution:
i) 2 × 103
= 2 × 10 × 10 × 10 = 2,000

ii) 72 × 22
= 7 × 7 × 2 × 2
= 49 × 4 = 196

KSEEB Solutions

iii) 23 × 5
= 2 × 2 × 2 × 5
= 8 × 5 = 40

iv) 3 × 44
= 3 × 4 × 4 × 4 × 4
= 12 × 64 = 768

v) 0 × 102
= 0 × 10 × 10 = 0

vi) 52 × 33
= 5 × 5 × 3 × 3 × 3
= 25 × 27 = 675

vii) 24 × 32
=2 × 2 × 2 × 2 × 3 × 3
= 16 × 9 = 144

viii) 32 × 104
= 3 × 3 × 10 × 10 × 10 × 10
= 90 × 1000 = 90,000

Question 7.
Simplify :
Solution:
i) (-4)3
(-4)3 = (-4) × (-4) × (-4) = -64

ii) (-3) × (-2)3
= (-3) × (-2) × (-2) × (-2) = (-3) × (-8) = 24

iii) (-3)2 × (-5)2
= (-3) × (-3) × (-5) × (-5)
= 9 × 25 = 225

iv) (-2)3 × (-10)3
= (-2) × (-2) × (-2) × (-10) × (-10) × (-10)
= +4 × -2 × 100 × -10
= -8 × -1000 = 8000

KSEEB Solutions

Question 8.
Compare the following numbers :
Solution:
i) 2 : 7 × 1012 ; 1.5 × 108
= 2.7 × 1012
= 2.7 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 2.70000000000000
= 27 × 1011

1.5 × 107
= 1.5 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 1.5 × 107
∴ 2.7 × 1012 >1.5 × 108

ii) 4 × 1014 ; 3 × 1017
=4 × 1014 = 4 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 4 × 1014
3 × 107 = 3 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
∴ 3 × 1017 > 4 × 1014

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