NCERT Solutions for Class 7 Maths Ch 13 Exponents and Powers
Exercise 13.1
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
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 x d
3. Express each of the following numbers using exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Answer
(i) 512
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
| 2 | 512 |
| 2 | 256 |
| 2 | 128 |
| 2 | 64 |
| 2 | 32 |
| 2 | 16 |
| 2 | 8 |
| 2 | 4 |
| 2 | 2 |
| 1 |
(ii) 343
343 = 7 × 7 × 7 = 73
| 7 | 343 |
| 7 | 49 |
| 7 | 7 |
| 1 |
(iii) 729
729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
| 3 | 729 |
| 3 | 243 |
| 3 | 81 |
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
(iv) 3125
3125 = 5 × 5 × 5 × 5 × 5 = 55
| 5 | 3125 |
| 5 | 625 |
| 5 | 125 |
| 5 | 25 |
| 5 | 5 |
| 1 |
4. Identify the greater number, wherever possible, in each of the following:
(i) 43 and 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
Since 64 < 81
Thus, 34 is greater than 43.
(ii) 53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
Since, 125 < 243
Thus, 34 is greater than 53.
(iii) 28= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8×8 = 64
Since, 256 > 64
Thus, 28 is greater than 82.
(iv) 1002 = 100 × 100 = 10,000
2100 = 2 × 2 × 2 × 2 × 2 .......14 times ....× 2 = 16,384 ....× 2
Since, 10,000 < 16,384 ....×2
Thus, 2100 is greater than 1002.
(v) 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1,024
102 = 10×10 = 100
Since, 1,024 > 100
Thus, 210>102
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 = 23× 34
| 2 | 648 |
| 2 | 324 |
| 2 | 162 |
| 3 | 81 |
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
(ii) 405 = 5 × 34
| 5 | 405 |
| 3 | 81 |
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
(iii) 540 = 22× 33× 5
| 2 | 540 |
| 2 | 270 |
| 3 | 135 |
| 3 | 45 |
| 3 | 15 |
| 5 | 5 |
| 1 |
(iv) 3,600 = 24× 32× 52
| 2 | 3600 |
| 2 | 1800 |
| 2 | 900 |
| 2 | 450 |
| 3 | 225 |
| 3 | 75 |
| 5 | 25 |
| 5 | 5 |
| 1 |
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 = 2,000
(ii) 72× 22 = 7 × 7 × 2 × 2 = 196
(iii) 23× 5 = 2 × 2 × 2 × 5 = 40
(iv) 3 × 44 = 3 × 4 × 4 × 4 × 4 = 768
(v) 0 × 102 = 0 × 10 × 10 = 0
(vi) 53× 33 = 5 × 5 × 3 × 3 × 3 = 675
(vii) 24× 32 = 2 × 2 × 2 × 2 × 3 × 3 = 144
(viii) 32× 104 = 3 × 3 × 10 × 10 × 10 × 10 = 90,000
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) = 225
(iv) (-2)3×(-10)3 =(-2)×(-2)×(-2)×(-10)×(-10)×(-10)
8. Compare the following numbers:
(i) 2.7 × 1012; 1.5 × 108
(ii) 4 × 1014; 3 × 1017
Answer
(i) 2.7 × 1012 and 1.5 × 108
On comparing the exponents of base 10,
2.7 × 1012> 1.5 × 108
(ii) 4 × 1014 and 3 × 1017
On comparing the exponents of base 10,
4 × 1014< 3 × 1017
Exercise 13.2
1. Using laws of exponents, simplify and write the answer in exponential form:
(i) 32 × 34 x 38
(ii) 615÷ 610
(iii) a3 × a2
(iv) 7 × x 72
(v) (52)2÷ 53
(vi) 25 × 55
(vii) a4 × b4
(viii) (34)3
(ix) [220÷ 215) × 23
(x) 8t × 82
Answer
(i) 32× 34× 38 = 3(2+4+8) = 314 [∵ am× an = am+n]
(ii) 615÷ 610 = 615-10 = 65 [∵ am÷ an = am-n]
(iii) a3 × a2 = a3+2 = a5 [∵ am × an = a m+n]
(iv) 7x× 72 = 7x+2 [∵ am× an = am+n]
(v) (52)2 ÷ 53 = 52×3÷ 53 = 56÷ 53 [∵ (am)n = am×n]
= 56-3 = 53 [∵ am÷ an = am-n]
(vi) 25× 25 = (2×5)5= 105 [∵ am×bm = (a×b)m]
(vii) a4× b4 = (a×b)4 [∵ am×bm = (a×b)m]
(viii) (34)3 = 34×3 = 312 [∵ (am)n = am×n]
(ix) (220 ÷ 215)×23 = (220-15)×23 [∵ am÷ an = am-n]
= 25× 23 = 25+3 = 28 [∵ am× an = am+n]
(x) 8t÷ 82 = 8t-2 [∵ am÷ an = am-n]
2. Simplify and express each of the following in exponential form:
(i) 23× 34× 4/3×32
(ii) [(52)3×54] ÷ 57
(iii) 254÷ 53
(iv) 3 × 72× 118 / 21 × 11
(v) 37 / 34× 33
(vi) 20 + 30 + 40
(vii) 20× 30× 40
(viii) (30 + 20) × 50
(ix) 28× a 5 / 43× a3
(x) (a5 / a3) × a8
(xi) 45× a8b3 / 45× a5b2
(xii) (23× 2)3
Answer
(i) 23× 34× 4 / 3× 32 = 23× 34× 22 / 3 × 25 = 23+2× 34 / 3×25 [∵ am× an = am+n]
= 25× 34 / 3 × 25 = 25-5× 34-3 [∵ am ÷ an = am-n]
= 20× 33 = 1× 33 = 33
(ii) [(52)3× 54] ÷ 57 [∵ (am)n = am×n]
= [56+4] ÷ 57 = 510÷ 57 [∵ am× an = am+n]
= 510-7 = 53 [∵ am ÷ an = am-n]
(iii) 254÷ 53 = (52)4÷ 53 = 58÷ 53 [∵ (am)n = am×n]
= 58-3 = 55 [∵ am ÷ an = am-n]
(iv) 3 × 72× 118 / 21 × 112 = 3 × 72 × 118 / 3 × 7 × 113 = 31-1× 72-1× 118-3 [∵ am ÷ an = am-n]
= 30× 71× 115 = 7 × 115
(v) 37 / 34× 33 = 37 / × 34+3 = 37 / 37 [∵ am× an = am+n]
= 37-7 = 30 = 1 [∵ am× an = am+n]
(vi) 20 + 30 + 40 + 1+1+1 = 3 [∵ a0 = 1]
(vii) 20× 30× 40 = 1×1×1 = 1 [∵ a0 = 1]
(viii) (30 + 20) × 50 = (1+1)×1 = 2×1 = 2 [∵ a0 = 1]
(ix) 28× a5 / 43× a3 = 28× a5/ (22)3× a3 = 28×a5 / 26× a3 [∵ (am)n = am×n]
= 28-6× a5-2 = 22× a2 [∵ am ÷ an = am-n]
= (2a)2 [∵ am×bm = (a×b)m]
(x) (a5 / a3) × a8 = (a 5-3) × a8 = a2× a8 [∵ am ÷ an = am-n]
= a2+8 = a10 [∵ am× an = am+n]
(xi) 45× a8b3 / 45× a5b2 = 45-5× a8-5× b3-2 = 40× a3× b [∵ am ÷ an = am-n]
= 1 × a3× b = a3b [∵ a0 = 1]
(xii) (23× 2)2 = (23+1)2 = (24)2 [∵ am× an = am+n]
= 24×2 = 28
3. Say true or false and justify your answer:
(i) 10 x 1011 = 10011
(ii) 23> 52
(iii) 23× 32 = 6s
(iv) 30 = (1000)0
Answer
(i) 10×1011 = 10011
L.H.S. 101+11 = 1012 and R.H.S. (102)11=1022
Since, L.H.S. ≠ R.H.S.
Therefore, it is false.
(ii) 23> 52
L.H.S 23 = 8 and R.H.S. 52 = 25
Since, L.H,S. is not greater than R.H.S.
Therefore, it is false,
(iii) 23×32 = 65
L.H.S. 23 × 32 = 8 x 9 = 72 and R.H.S. 65 =7.776
Since, L.H.S. ≠ R.H.S.
Therefore, it is false.
(iv) 30 = (1000)0
L.H.S. 30 = 1 and R.H.S. (1000)0 = 1
Since, L.H.S. = R.H.S.
Therefore, it is true.
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)
= 36× 25
(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 28× 3
5. Simplify:
(i) (25)2× 73 / 83× 7
(ii) 25 × 52× t8 / 103× t4
(iii) 35× 105× 25 × 57× 65
Answer
(i) (25)2× 73 / 83× 7 = 25×2× 73 / (23)3× 7
= 210× 73 / 29× 7
= 210-9× 73-1
= 2 × 72
= 2 × 49
= 98
(ii) 25 × 52×t8 / 103 - t4 = 52× 52× t8 / (5×2)3× t4
= 52+2× t8-4/23×33
= 54× t4 / 23× 53
= 54-3× t4 / 23
= 5t4 / 8
(iii) 35× 105× 25 / 57× 65= 35×(2×5)5× 52 / 57× (2×3)5
= 35× 25× 55× 52 / 57× 25× 35
= 35× 25× 55+2 / 57× 25× 35
= 35× 25× 57 / 57× 25× 35
= 25-5× 35-5× 55-5
= 20× 30× 50
= 1 × 1 × 1
= 1
Exercise 13.3
1. Write the following numbers in the expanded form:
(i) 279404
(ii) 3006194
(iii) 2806196
(iv) 120719
(v) 20068
Answer
(i) 2,79,404
= 2,00,000 + 70,000 + 9,000 + 400 + 00 + 4
= (2×100000) + (7 × 10000) + (9×1000) + (4 × 100) + (0× 10) + (4×1)
= (2×105) + (7×104) + (9×103) + (4 ×102) + (0×101) + (4×100)
(ii) 3006194
= 3000000 + 0 + 0 + 6000 + 100 + 90 + 4
= (3 × 1000000) + (0 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + 4
= (3 × 106) + (0 × 105) + (0 × 104) + (6 × 103) + (1 × 102) + (9 × 101) + (4 × 100)
(iii) 28,06,196
= 20,00,000 + 8,00,000 + 0 + 6,000 + 100 + 90 + 6= 2×1000000 + 8× 100000+ 0 × 10000 +6 × 1000 + 1 × 100 + 9 × 10 + 6 × 1
= 2 ×106+8×105+0×104 + 6×103+1x102+9×10+6×100
(iv) 1,20,719 = 1,00,000 + 20,000 + 0 + 700 + 10 + 9
= 1 × 100000 + 2 × 10000 + 0× 1000 + 7×100 + 1× 10+9×1
= 1 × 105 + 2 × 104 + 0 x 103 + 7 × 102 + 1 x 101 + 9 × 100
= 1 × 100000 + 2 × 10000 + 0× 1000 + 7×100 + 1× 10+9×1
= 1 × 105 + 2 × 104 + 0 x 103 + 7 × 102 + 1 x 101 + 9 × 100
(v) 20,068= 20,000 + 00 + 00 + 60 + 8
= 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8 × 1
= 2×104 + 0 × 103 + 0 × 102 + 6 x 101 + 8 × 100
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 × 103 + 7 × 102 + 5 × 100
(d) 9 × 105 + 2 × 102 + 3 × 101
Answer
(a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
= 8 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1
= 80000 + 6000 + 0 + 40+5
= 86,045
(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
= 4 × 100000 + 0 × 10000 + 5 × 1000 + 3 × 100 + 0 × 10 + 2 × 1
= 400000 + 0 + 5000 + 3000 + 0 + 2
= 4,05,302
(c) 3 × 104 + 7 × 102 + 5 × 100
= 3 × 10000 + 0 × 1000 + 7 × 100 + 0 × 10 + 5 × 1
= 30000 + 0 + 700 + 0 + 5
= 30,705
(d) 9 × 105 + 2 × 102 + 3 × 101
= 9 × 100000 + 0 × 10000 + 0 × 1000 + 2 × 100 + 3 × 10 + 0 × 1
= 900000 + 0 + 0 + 200 + 30 + 0
= 9,00,230
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) 5,00,00,000 = 5 × 1,00,00,000 = 5 × 107
(ii) 70,00,000 = 7 × 10,00,000 = 7 × 106
(iii) 3,18,65,00,000 = 31865 × 100000
= 3.1865 × 10000 × 100000 = 3.1865 × 109
(iv) 3,90,878 = 3.90878 × 100000 = 3.90878 × 105
(v) 39087.8 = 3.90878 × 10000 = 3.90878 × 104
(vi) 3908.78 = 3.90878 × 1000 = 3.90878 × 103
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 Earth id 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 100,000,000,0000 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) The distance between Earth and Moon = 384,000,000 m
= 384 × 1000000 m
= 3.84 × 100 × 1000000
= 3.84 × 108 m
(b) Speed of light in vacuum = 300,000,000 m/s
= 3 × 100000000 m/s
= 3×108 m/s
(c) Diameter of the Earth = 1,27,50,000 m
= 12756 × 1000 m
= 1.2756 × 10000 × 1000 m
= 1 2756×107 m
(d) Diameter of the Sun = 1,400,000,000 m
= 14 × 100,000,000 m
= 1.4 × 10 × 100,000,000 m
= 1.4 × 109 m
(e) Average of Stars = 100, 000, 000,000
= 1 × 100,000,000,000
= 1×1011
(f) Years of Universe = 12,000,000,000 years
= 12 × 1000,000,000 years
= 1.2 × 10 × 1000,000,000 years
= 1.2 × 1010 years
(g) Distance of the Sun from the centre of the Milky Way Galaxy
= 300,000,000,000,000,000,000 m
= 3 × 100,000,000,000,000,000,000 m
= 3 × 1020 m
(h) Number of molecules in a drop of water weighing 1.8 gm
= 60,230,000,000,000,000,000,000
= 6023 × 10,000,000,000,000,000,000
= 6,023 × 1000 × 10,000,000,000,000,000,000
= 6.023 × 1022
(i) The Earth has Sea water = 1,353,000,000 km3
= 1,353 × 1000000 km3
= 1,353 × 1000 × 1000,000 km3
= 1.353 × 109 km3
(j) The population of India = 1,027,000,000
= 1027 × 1000000 = 1,027 × 1000 × 1000000
= 1.027×109.