Students can Download Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
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 = 2 × 2 × 2 × 2 × 2 × 2 = 64
(ii) 93 = 9 × 9 × 9 = 729
(iii) 112= 11 × 11 = 121
(iv) 54 = 5 × 5 × 5 × 5 = 625
Question 2.
Express the following in exponent 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
Question 3.
Express each of the following numbers using the exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Solution:
(i) 512
Prime factorizing
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 29
∴ 512 = 29
(ii) 343
Prime factorizing
= 7 × 7 × 7
Hence, 343 = 73
(iii) 729
Prime factorizing
= 3 × 3 × 3 × 3 × 3 × 3
= 36
Hence, 729 = 36
(iv) 3125
Prime factorizing
= 5 × 5 × 5 × 5 × 5 × 5
= 55
Question 4.
Identify the greater number, wherever possible, in each of the following?
(i) 43 or 34
(ii) 53 or 35
(iii) 28 or 82
(iv) 1002 or 2100
(v) 210 or 102
Solution:
(i) 43 or 34
43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
81 > 64
Hence 34 > 43
(ii) 53 or 35
53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
243 > 125
Hence 35 > 53
(iii) 28 or 82
28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
256 > 64
Hence 28 > 82
(iv) 1002 or 2100
Clearly 2100
(v) 210 or 102
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
102 = 10 × 10= 100
1024 > 100
Hence 210 > 102
Question 5.
Express each of the following as product of powers of their prime factors:
(i) 648
(ii) 405
(iii) 540
(iv) 3,600
Solution:
(i) 648
Prime factorizing
2 × 2 × 2 × 3 × 3 × 3 × 3
648 = 23 × 34
(ii) 405
Prime factorizing
3 × 3 × 3 × 3 × 5
Hence, 405 = 34 × 5
(iii) 540
Prime factorizing
2 × 2 × 3 × 3 × 3 × 5
540 = 22 × 33 × 5
(iv) 3600
Prime factorizing
= 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
3600 = 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
103 = 10 × 10 × 10 = 1000
2 × 1000 = 2000
(ii) 72 × 22
72 = 7 × 7 = 49, 22 = 2 × 2=4
49 × 4=196
(iii) 23 × 5
23 = 2 × 2 × 2=8
8 × 5 = 40
(iv) 3 × 44
44 = 4 × 4 × 4 × 4 = 256
23 × 256 = 768
(v) 0 × 102 = 0
(vi) 52 × 33
5 × 5 × 3 × 3 × 3
25 × 27 = 675
(vi) 24 × 32
2 × 2 × 2 × 2 × 3 × 3
16×9= 144
(vii) 32 × 104
3 × 3 × 10 × 10 × 10 × 10
9 × 10000 = 90000
Question 7.
Simplify:
(i) (- 4)3
(ii) (- 3) × (- 2)3
(iii) (- 3)2 × (- 5)2
(iv) (- 2)3 × (- 10)3
Solution:
(i) (- 4)3
(- 4) × (- 4) × (- 4) = – 64
(ii) (- 3) × (- 2)3
(- 2)3 = (- 2) × (- 2) × (- 2) = – 8
(- 3) × (-8) = 24
(iii) (- 3)2 × (- 5)2
(-3)2 = (- 3) × (- 3) = 9
(- 5)2 = (- 5) × (- 5) = 25
9 × 25 = 225
(iv) (- 2)3 × (- 10)3
(- 2)3 = (- 2) × (- 2) × (- 2) = – 8
(- 10)3 = (- 10)(- 10)(- 10) = – 1000
– 8 × (- 1000) = 8000
Question 8.
Compare the following numbers:
(i) 2.7 × 1012, 1.5 × 108
(ii) 4 × 14 , 3 × 1017
Solution:
(i) 2.7 × 1012 = 2.7 × 1000000000000
= 2700000000000
1.5 × 108 = 1.5 × 100000000
= 150000000
2.7 × 102 > 1.5 × 108
(ii) 4 × 1014, 3 × 1017
4 × 1014 = 400000000000000
3 × 1017 = 300000000000000000
Hence, 4 × 1014 < 3 × 1017