NCERT Solutions for Class 7 Maths Chapter 13 – Exponents and Powers
Page No 252:
Question 1:
Find the value of:
(i) 26 (ii) 93
(iii) 112 (iv)54
Answer:
(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 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
Answer:
(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
Page No 253:
Question 3:
Express the following numbers using exponential notation:
(i) 512 (ii) 343
(iii) 729 (iv) 3125
Answer:
(i) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
(ii) 343 = 7 × 7 × 7 = 73
(iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
(iv) 3125 = 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
Answer:
(i) 43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
Therefore, 34 > 43
(ii) 53 = 5 × 5 × 5 =125
35 = 3 × 3 × 3 × 3 × 3 = 243
Therefore, 35 > 53
(iii) 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
Therefore, 28 > 82
(iv)1002 or 2100
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
2100 = 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 × 1024 ×1024 × 1024
1002 = 100 × 100 = 10000
Therefore, 2100 > 1002
(v) 210 and 102
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
102 = 10 × 10 = 100
Therefore, 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
Answer:
(i) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 23. 34
(ii) 405 = 3 × 3 × 3 × 3 × 5 = 34 . 5
(iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 22. 33. 5
(iv) 3600 = 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
Answer:
(i) 2 × 103 = 2 × 10 × 10 × 10 = 2 × 1000 = 2000
(ii) 72 × 22 = 7 × 7 × 2 × 2 = 49 × 4 = 196
(iii) 23 × 5 = 2 × 2 × 2 × 5 = 8 × 5 = 40
(iv) 3 × 44 = 3 × 4 × 4 × 4 × 4 = 3 × 256 = 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 = 9 × 10000 = 90000
Question 7:
Simplify:
(i) (− 4)3 (ii) (− 3) × (− 2)3
(iii) (− 3)2 × (− 5)2 (iv)(− 2)3 × (−10)3
Answer:
(i) (−4)3 = (−4) × (−4) × (−4) = −64
(ii) (−3) × (−2)3 = (−3) × (−2) × (−2) × (−2) = 24
(iii) (−3)2 × (−5)2 = (−3) × (−3) × (−5) × (−5) = 9 × 25 = 225
(iv) (−2)3 × (−10)3 = (−2) × (−2) × (−2) × (−10) × (−10) × (−10)
= (−8) × (−1000) = 8000
Question 8:
Compare the following numbers:
(i) 2.7 × 1012; 1.5 × 108
(ii) 4 × 1014; 3 × 1017
Answer:
(i) 2.7 × 1012; 1.5 × 108
2.7 × 1012 > 1.5 × 108
(ii) 4 × 1014; 3 × 1017
3 × 1017 > 4 × 1014
Page No 260:
Question 1:
Using laws of exponents, simplify and write the answer in exponential form:
(i) 32 × 34 × 38 (ii) 615 ÷ 610 (iii) a3 × a2
(iv) 7x× 72 (v) 
(vi) 25 × 55
(vi) 25 × 55
(vii) a4 × b4 (viii) (34)3
(ix) 
(x) 8t ÷ 82
(x) 8t ÷ 82Answer:
(i) 32 × 34 × 38 = (3)2 + 4 + 8 (am × an = am+n)
= 314
(ii) 615 ÷ 610 = (6)15 − 10 (am ÷ an = am−n)
= 65
(iii) a3 × a2 = a(3 + 2) (am × an = am+n)
= a5
(iv) 7x + 72 = 7x + 2 (am × an = am+n)
(v) (52)3 ÷ 53
= 52 × 3 ÷ 53 (am)n = amn
= 56 ÷ 53
= 5(6 − 3) (am ÷ an = am−n)
= 53
(vi) 25 × 55
= (2 × 5)5 [am × bm = (a × b)m]
= 105
(vii) a4 × b4
= (ab)4 [am × bm = (a × b)m]
(viii) (34)3 = 34 × 3 = 312 (am)n = amn
(ix) (220 ÷ 215) × 23
= (220 − 15) × 23 (am ÷ an = am−n)
= 25 × 23
= (25 + 3) (am × an = am+n)
= 28
(x) 8t ÷ 82 = 8(t − 2) (am ÷ an = am−n)
Question 2:
Simplify and express each of the following in exponential form:
(i) 
(ii) 
(iii) 
(ii) (iii)
(iv) 
(v) 
(vi) 20 + 30 + 40
(v) (vi) 20 + 30 + 40
(vii) 20 × 30 × 40 (viii) (30 + 20) × 50 (ix) 
(x) 
(xi) 
(xii) 
(xi) (xii) Answer:
(i)
(ii) [(52)3 × 54] ÷ 57
= [52 × 3 × 54] ÷ 57 (am)n = amn
= [56 × 54] ÷ 57
= [56 + 4] ÷ 57 (am × an = am+n)
= 510 ÷ 57
= 510 − 7 (am ÷ an = am−n)
= 53
(iii) 254 ÷ 53 = (5 ×5)4 ÷ 53
= (52)4 ÷ 53
= 52 × 4 ÷ 53 (am)n = amn
= 58 ÷ 53
= 58 − 3 (am ÷ an = am−n)
= 55
(iv)
= 1 × 7 × 115 = 7 × 115
(v)
(vi) 20 + 30 + 40 = 1 + 1 + 1 = 3
(vii) 20 × 30 × 40 = 1 × 1 × 1 = 1
(viii) (30 + 20) × 50 = (1 + 1) × 1 = 2
(ix)
(x)
(xi)
(xii) (23 × 2)2 = 
(am × an = am+n)
(am × an = am+n)
= (24)2 = 24 × 2 (am)n = amn
= 28
Question 3:
Say true or false and justify your answer:
(i) 10 × 1011 = 10011 (ii) 23 > 52
(iii) 23 × 32 = 65 (iv) 30 = (1000)0
Answer:
(i) 10 × 1011 = 10011
L.H.S. = 10 × 1011 = 1011 + 1 (am × an = am+n)
= 1012
R.H.S. = 10011 = (10 ×10)11= (102)11
= 102 × 11 = 1022 (am)n = amn
As L.H.S. ≠ R.H.S.,
Therefore, the given statement is false.
(ii) 23 > 52
L.H.S. = 23 = 2 × 2 × 2 = 8
R.H.S. = 52 = 5 × 5 = 25
As 25 > 8,
Therefore, the given statement is false.
(iii) 23 × 32 = 65
L.H.S. = 23 × 32 = 2 × 2 × 2 × 3 × 3 = 72
R.H.S. = 65 = 7776
As L.H.S. ≠ R.H.S.,
Therefore, the given statement is false.
(iv) 30 = (1000)0
L.H.S. = 30 = 1
R.H.S. = (1000)0 = 1 = L.H.S.
Therefore, the given statement is true.
Page No 261:
Question 4:
Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192 (ii) 270
(iii) 729 × 64 (iv) 768
Answer:
(i) 108 × 192
= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)
= (22 × 33) × (26 × 3)
= 26 + 2 × 33 + 1 (am × an = am+n)
= 28 × 34
(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 33 × 5
(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2)
= 36 × 26
(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 28 × 3
Question 5:
Simplify:
(i) 
(ii) 
(iii) 
(ii) (iii) Answer:
(i)
(ii)
(iii)
Page No 263:
Question 1:
Write the following numbers in the expanded forms:
279404, 3006194, 2806196, 120719, 20068
Answer:
279404 = 2 × 105 + 7 × 104 + 9 × 103 + 4 × 102 + 0 × 101 + 4 × 100
3006194 = 3 × 106 + 0 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 101 + 4 × 100
2806196 = 2 × 106 + 8 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 101 + 6 × 100
120719 = 1 × 105 + 2 × 104 + 0 × 103 + 7 × 102 + 1 × 101 + 9 × 100
20068 = 2 × 104 + 0 × 103 + 0 × 102 + 6 × 101 + 8 × 100
Question 2:
Find the number from each of the following expanded forms:
(a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
(c) 3 × 104 + 7 × 102 + 5 × 100
(d) 9 × 105 + 2 × 102 + 3 × 101
Answer:
(a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
= 86045
(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
= 405302
(c) 3 × 104 + 7 × 102 + 5 × 100
= 30705
(d) 9 × 105 + 2 × 102 + 3 × 101
= 900230
Question 3:
Express the following numbers in standard form:
(i) 5, 00, 00, 000 (ii) 70, 00, 000
(iii) 3, 18, 65, 00, 000 (iv) 3, 90, 878
(v) 39087.8 (vi) 3908.78
Answer:
(i) 50000000 = 5 × 107
(ii) 7000000 = 7 × 106
(iii) 3186500000 = 3.1865 × 109
(iv) 390878 = 3.90878 × 105
(v) 39087.8 = 3.90878 × 104
(vi) 3908.78 = 3.90878 × 103
Question 4:
Express the number appearing in the following statements in standard form.
(a) The distance between Earth and Moon is 384, 000, 000 m.
(b) Speed of light in vacuum is 300, 000, 000 m/s.
(c) Diameter of the Earth is 1, 27, 56, 000 m.
(d) Diameter of the Sun is 1, 400, 000, 000 m.
(e) In a galaxy there are on an average 100, 000, 000, 000 stars.
(f) The universe is estimated to be about 12, 000, 000, 000 years old.
(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300, 000, 000, 000, 000, 000, 000 m.
(h) 60, 230, 000, 000, 000, 000, 000, 000 molecules are contained in a drop of water weighing 1.8 gm.
(i) The earth has 1, 353, 000, 000 cubic km of sea water.
(j) The population of India was about 1, 027, 000, 000 in March, 2001.
Answer:
(a) 3.84 × 108 m
(b) 3 × 108 m/s
(c) 1.2756 × 107 m
(d) 1.4 × 109 m
(e) 1 × 1011 stars
(f) 1.2 × 1010 years
(g) 3 × 1020 m
(h) 6.023 × 1022
(i) 1.353 × 109 cubic km
(j) 1.027 × 109
