NCERT Solutions for Class 7 Maths Chapter 11 Exponents and Powers
Exercise 11.1
1. Find the value of:
(i) 26
Solution:-
The above value can be written as,
= 2 × 2 × 2 × 2 × 2 × 2
= 64
(ii) 93
Solution:-
The above value can be written as,
= 9 × 9 × 9
= 729
(iii) 112
Solution:-
The above value can be written as,
= 11 × 11
= 121
(iv) 54
Solution:-
The above value can be written as,
= 5 × 5 × 5 × 5
= 625
2. Express the following in exponential form:
(i) 6 × 6 × 6 × 6
Solution:-
The given question can be expressed in the exponential form as 64.
(ii) t × t
Solution:-
The given question can be expressed in the exponential form as t2.
(iii) b × b × b × b
Solution:-
The given question can be expressed in the exponential form as b4.
(iv) 5 × 5× 7 × 7 × 7
Solution:-
The given question can be expressed in the exponential form as 52 × 73.
(v) 2 × 2 × a × a
Solution:-
The given question can be expressed in the exponential form as 22 × a2.
(vi) a × a × a × c × c × c × c × d
Solution:-
The given question can be expressed in the exponential form as a3 × c4 × d.
3. Express each of the following numbers using the exponential notation:
(i) 512
Solution:-
The factors of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
So it can be expressed in the exponential form as 29.
(ii) 343
Solution:-
The factors of 343 = 7 × 7 × 7
So it can be expressed in the exponential form as 73.
(iii) 729
Solution:-
The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3
So it can be expressed in the exponential form as 36.
(iv) 3125
Solution:-
The factors of 3125 = 5 × 5 × 5 × 5 × 5
So it can be expressed in the exponential form as 55.
4. Identify the greater number, wherever possible, in each of the following.
(i) 43 or 34
Solution:-
The expansion of 43 = 4 × 4 × 4 = 64
The expansion of 34 = 3 × 3 × 3 × 3 = 81
Clearly,
64 < 81
So, 43 < 34
Hence, 34 is the greater number.
(ii) 53 or 35
Solution:-
The expansion of 53 = 5 × 5 × 5 = 125
The expansion of 35 = 3 × 3 × 3 × 3 × 3= 243
Clearly,
125 < 243
So, 53 < 35
Hence, 35 is the greater number.
(iii) 28 or 82
Solution:-
The expansion of 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
The expansion of 82 = 8 × 8= 64
Clearly,
256 > 64
So, 28 > 82
Hence, 28 is the greater number.
(iv) 1002 or 2100
Solution:-
The expansion of 1002 = 100 × 100 = 10000
The expansion of 2100
210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
Then,
2100 = 1024 × 1024 ×1024 × 1024 ×1024 × 1024 × 1024 × 1024 × 1024 × 1024 = (1024)10
Clearly,
1002 < 2100
Hence, 2100 is the greater number.
(v) 210 or 102
Solution:-
The expansion of 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
The expansion of 102 = 10 × 10= 100
Clearly,
1024 > 100
So, 210 > 102
Hence, 210 is the greater number.
5. Express each of the following as a product of powers of their prime factors:
(i) 648
Solution:-
Factors of 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3
= 23 × 34
(ii) 405
Solution:-
Factors of 405 = 3 × 3 × 3 × 3 × 5
= 34 × 5
(iii) 540
Solution:-
Factors of 540 = 2 × 2 × 3 × 3 × 3 × 5
= 22 × 33 × 5
(iv) 3,600
Solution:-
Factors of 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 24 × 32 × 52
6. Simplify:
(i) 2 × 103
Solution:-
The above question can be written as,
= 2 × 10 × 10 × 10
= 2 × 1000
= 2000
(ii) 72 × 22
Solution:-
The above question can be written as,
= 7 × 7 × 2 × 2
= 49 × 4
= 196
(iii) 23 × 5
Solution:-
The above question can be written as,
= 2 × 2 × 2 × 5
= 8 × 5
= 40
(iv) 3 × 44
Solution:-
The above question can be written as,
= 3 × 4 × 4 × 4 × 4
= 3 × 256
= 768
(v) 0 × 102
Solution:-
The above question can be written as,
= 0 × 10 × 10
= 0 × 100
= 0
(vi) 52 × 33
Solution:-
The above question can be written as,
= 5 × 5 × 3 × 3 × 3
= 25 × 27
= 675
(vii) 24 × 32
Solution:-
The above question can be written as,
= 2 × 2 × 2 × 2 × 3 × 3
= 16 × 9
= 144
(viii) 32 × 104
Solution:-
The above question can be written as,
= 3 × 3 × 10 × 10 × 10 × 10
= 9 × 10000
= 90000
7. Simplify:
(i) (– 4)3
Solution:-
The expansion of -43
= – 4 × – 4 × – 4
= – 64
(ii) (–3) × (–2)3
Solution:-
The expansion of (-3) × (-2)3
= – 3 × – 2 × – 2 × – 2
= – 3 × – 8
= 24
(iii) (–3)2 × (–5)2
Solution:-
The expansion of (-3)2 × (-5)2
= – 3 × – 3 × – 5 × – 5
= 9 × 25
= 225
(iv) (–2)3 × (–10)3
Solution:-
The expansion of (-2)3 × (-10)3
= – 2 × – 2 × – 2 × – 10 × – 10 × – 10
= – 8 × – 1000
= 8000
8. Compare the following numbers:
(i) 2.7 × 1012 ; 1.5 × 108
Solution:-
By observing the question
Comparing the exponents of base 10,
Clearly,
2.7 × 1012 > 1.5 × 108
(ii) 4 × 1014 ; 3 × 1017
Solution:-
By observing the question
Comparing the exponents of base 10,
Clearly,
4 × 1014 < 3 × 1017
Exercise 11.2
1. Using laws of exponents, simplify and write the answer in exponential form:
(i) 32 × 34 × 38
Solution:-
By the rule of multiplying the powers with the same base = am × an = am + n
Then,
= (3)2 + 4 + 8
= 314
(ii) 615 ÷ 610
Solution:-
By the rule of dividing the powers with the same base = am ÷ an = am – n
Then,
= (6)15 – 10
= 65
(iii) a3 × a2
Solution:-
By the rule of multiplying the powers with the same base = am × an = am + n
Then,
= (a)3 + 2
= a5
(iv) 7x × 72
Solution:-
By the rule of multiplying the powers with the same base = am × an = am + n
Then,
= (7)x + 2
(v) (52)3 ÷ 53
Solution:-
By the rule of taking the power of as power = (am)n = amn
(52)3 can be written as = (5)2 × 3
= 56
Now, 56 ÷ 53
By the rule of dividing the powers with the same base = am ÷ an = am – n
Then,
= (5)6 – 3
= 53
(vi) 25 × 55
Solution:-
By the rule of multiplying the powers with the same exponents = am × bm = abm
Then,
= (2 × 5)5
= 105
(vii) a4 × b4
Solution:-
By the rule of multiplying the powers with the same exponents = am × bm = abm
Then,
= (a × b)4
= ab4
(viii) (34)3
Solution:-
By the rule of taking the power of as power = (am)n = amn
(34)3 can be written as = (3)4 × 3
= 312
(ix) (220 ÷ 215) × 23
Solution:-
By the rule of dividing the powers with the same base = am ÷ an = am – n
(220 ÷ 215) can be simplified as,
= (2)20 – 15
= 25
Then,
By the rule of multiplying the powers with the same base = am × an = am + n
25 × 23 can be simplified as,
= (2)5 + 3
= 28
(x) 8t ÷ 82
Solution:-
By the rule of dividing the powers with the same base = am ÷ an = am – n
Then,
= (8)t – 2
2. Simplify and express each of the following in exponential form:
(i) (23 × 34 × 4)/ (3 × 32)
Solution:-
Factors of 32 = 2 × 2 × 2 × 2 × 2
= 25
Factors of 4 = 2 × 2
= 22
Then,
= (23 × 34 × 22)/ (3 × 25)
= (23 + 2 × 34) / (3 × 25) … [∵am × an = am + n]
= (25 × 34) / (3 × 25)
= 25 – 5 × 34 – 1 … [∵am ÷ an = am – n]
= 20 × 33
= 1 × 33
= 33
(ii) ((52)3 × 54) ÷ 57
Solution:-
(52)3 can be written as = (5)2 × 3 … [∵(am)n = amn]
= 56
Then,
= (56 × 54) ÷ 57
= (56 + 4) ÷ 57 … [∵am × an = am + n]
= 510 ÷ 57
= 510 – 7 … [∵am ÷ an = am – n]
= 53
(iii) 254 ÷ 53
Solution:-
(25)4 can be written as = (5 × 5)4
= (52)4
(52)4 can be written as = (5)2 × 4 … [∵(am)n = amn]
= 58
Then,
= 58 ÷ 53
= 58 – 3 … [∵am ÷ an = am – n]
= 55
(iv) (3 × 72 × 118)/ (21 × 113)
Solution:-
Factors of 21 = 7 × 3
Then,
= (3 × 72 × 118)/ (7 × 3 × 113)
= 31-1 × 72-1 × 118 – 3
= 30 × 7 × 115
= 1 × 7 × 115
= 7 × 115
(v) 37/ (34 × 33)
Solution:-
= 37/ (34+3) … [∵am × an = am + n]
= 37/ 37
= 37 – 7 … [∵am ÷ an = am – n]
= 30
= 1
(vi) 20 + 30 + 40
Solution:-
= 1 + 1 + 1
= 3
(vii) 20 × 30 × 40
Solution:-
= 1 × 1 × 1
= 1
(viii) (30 + 20) × 50
Solution:-
= (1 + 1) × 1
= (2) × 1
= 2
(ix) (28 × a5)/ (43 × a3)
Solution:-
(4)3 can be written as = (2 × 2)3
= (22)3
(22)3 can be written as = (2)2 × 3 … [∵(am)n = amn]
= 26
Then,
= (28 × a5)/ (26 × a3)
= 28 – 6 × a5 – 3 … [∵am ÷ an = am – n]
= 22 × a2 … [∵(am)n = amn]
= 2a2
(x) (a5/a3) × a8
Solution:-
= (a5 -3) × a8 … [∵am ÷ an = am – n]
= a2 × a8
= a2 + 8 … [∵am × an = am + n]
= a10
(xi) (45 × a8b3)/ (45 × a5b2)
Solution:-
= 45 – 5 × (a8 – 5 × b3 – 2) … [∵am ÷ an = am – n]
= 40 × (a3b)
= 1 × a3b
= a3b
(xii) (23 × 2)2
Solution:-
= (23 + 1)2 … [∵am × an = am + n]
= (24)2
(24)2 can be written as = (2)4 × 2 … [∵(am)n = amn]
= 28
3. Say true or false and justify your answer:
(i) 10 × 1011 = 10011
Solution:-
Let us consider Left Hand Side (LHS) = 10 × 1011
= 101 + 11 … [∵am × an = am + n]
= 1012
Now, consider Right Hand Side (RHS) = 10011
= (10 × 10)11
= (101 + 1)11
= (102)11
= (10)2 × 11 … [∵(am)n = amn]
= 1022
By comparing LHS and RHS,
LHS ≠ RHS
Hence, the given statement is false.
(ii) 23 > 52
Solution:-
Let us consider LHS = 23
Expansion of 23 = 2 × 2 × 2
= 8
Now, consider RHS = 52
Expansion of 52 = 5 × 5
= 25
By comparing LHS and RHS,
LHS < RHS
23 < 52
Hence, the given statement is false.
(iii) 23 × 32 = 65
Solution:-
Let us consider LHS = 23 × 32
Expansion of 23 × 32= 2 × 2 × 2 × 3 × 3
= 72
Now, consider RHS = 65
Expansion of 65 = 6 × 6 × 6 × 6 × 6
= 7776
By comparing LHS and RHS,
72 ≠ 7776
LHS ≠ RHS
Hence, the given statement is false.
(iv) 30 = (1000)0
Solution:-
Let us consider LHS = 30
= 1
Now, consider RHS = 10000
= 1
By comparing LHS and RHS,
LHS = RHS
30 = 10000
Hence, the given statement is true.
4. Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
Solution:-
The factors of 108 = 2 × 2 × 3 × 3 × 3
= 22 × 33
The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
= 26 × 3
Then,
= (22 × 33) × (26 × 3)
= 22 + 6 × 33 + 1 … [∵am × an = am + n]
= 28 × 34
(ii) 270
Solution:-
The factors of 270 = 2 × 3 × 3 × 3 × 5
= 2 × 33 × 5
(iii) 729 × 64
The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3
= 36
The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2
= 26
Then,
= (36 × 26)
= 36 × 26
(iv) 768
Solution:-
The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
= 28 × 3
5. Simplify:
(i) ((25)2 × 73)/ (83 × 7)
Solution:-
83 can be written as = (2 × 2 × 2)3
= (23)3
We have,
= ((25)2 × 73)/ ((23)3 × 7)
= (25 × 2 × 73)/ ((23 × 3 × 7) … [∵(am)n = amn]
= (210 × 73)/ (29 × 7)
= (210 – 9 × 73 – 1) … [∵am ÷ an = am – n]
= 2 × 72
= 2 × 7 × 7
= 98
(ii) (25 × 52 × t8)/ (103 × t4)
Solution:-
25 can be written as = 5 × 5
= 52
103 can be written as = 103
= (5 × 2)3
= 53 × 23
We have,
= (52 × 52 × t8)/ (53 × 23 × t4)
= (52 + 2 × t8)/ (53 × 23 × t4) … [∵am × an = am + n]
= (54 × t8)/ (53 × 23 × t4)
= (54 – 3 × t8 – 4)/ 23 … [∵am ÷ an = am – n]
= (5 × t4)/ (2 × 2 × 2)
= (5t4)/ 8
(iii) (35 × 105 × 25)/ (57 × 65)
Solution:-
105 can be written as = (5 × 2)5
= 55 × 25
25 can be written as = 5 × 5
= 52
65 can be written as = (2 × 3)5
= 25 × 35
Then we have,
= (35 × 55 × 25 × 52)/ (57 × 25 × 35)
= (35 × 55 + 2 × 25)/ (57 × 25 × 35) … [∵am × an = am + n]
= (35 × 57 × 25)/ (57 × 25 × 35)
= (35 – 5 × 57 – 7 × 25 – 5)
= (30 × 50 × 20) … [∵am ÷ an = am – n]
= 1 × 1 × 1
= 1
Exercise 11.3
1. Write the following numbers in the expanded forms:
(a) 279404
Solution:-
The expanded form of the number 279404 is,
= (2 × 100000) + (7 × 10000) + (9 × 1000) + (4 × 100) + (0 × 10) + (4 × 1)
Now we can express it using powers of 10 in the exponent form,
= (2 × 105) + (7 × 104) + (9 × 103) + (4 × 102) + (0 × 101) + (4 × 100)
(b) 3006194
Solution:-
The expanded form of the number 3006194 is,
= (3 × 1000000) + (0 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + (4 × 1)
Now we can express it using powers of 10 in the exponent form,
= (3 × 106) + (0 × 105) + (0 × 104) + (6 × 103) + (1 × 102) + (9 × 101) + (4 × 100)
(c) 2806196
Solution:-
The expanded form of the number 2806196 is,
= (2 × 1000000) + (8 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + (6 × 1)
Now we can express it using powers of 10 in the exponent form,
= (2 × 106) + (8 × 105) + (0 × 104) + (6 × 103) + (1 × 102) + (9 × 101) + (6 × 100)
(d) 120719
Solution:-
The expanded form of the number 120719 is,
= (1 × 100000) + (2 × 10000) + (0 × 1000) + (7 × 100) + (1 × 10) + (9 × 1)
Now we can express it using powers of 10 in the exponent form,
= (1 × 105) + (2 × 104) + (0 × 103) + (7 × 102) + (1 × 101) + (9 × 100)
(e) 20068
Solution:-
The expanded form of the number 20068 is,
= (2 × 10000) + (0 × 1000) + (0 × 100) + (6 × 10) + (8 × 1)
Now we can express it using powers of 10 in the exponent form,
= (2 × 104) + (0 × 103) + (0 × 102) + (6 × 101) + (8 × 100)
2. Find the number from each of the following expanded forms:
(a) (8 × 10)4 + (6 × 10)3 + (0 × 10)2 + (4 × 10)1 + (5 × 10)0
Solution:-
The expanded form is,
= (8 × 10000) + (6 × 1000) + (0 × 100) + (4 × 10) + (5 × 1)
= 80000 + 6000 + 0 + 40 + 5
= 86045
(b) (4 × 10)5 + (5 × 10)3 + (3 × 10)2 + (2 × 10)0
Solution:-
The expanded form is,
= (4 × 100000) + (0 × 10000) + (5 × 1000) + (3 × 100) + (0 × 10) + (2 × 1)
= 400000 + 0 + 5000 + 300 + 0 + 2
= 405302
(c) (3 × 10)4 + (7 × 10)2 + (5 × 10)0
Solution:-
The expanded form is,
= (3 × 10000) + (0 × 1000) + (7 × 100) + (0 × 10) + (5 × 1)
= 30000 + 0 + 700 + 0 + 5
= 30705
(d) (9 × 10)5 + (2 × 10)2 + (3 × 10)1
Solution:-
The expanded form is,
= (9 × 100000) + (0 × 10000) + (0 × 1000) + (2 × 100) + (3 × 10) + (0 × 1)
= 900000 + 0 + 0 + 200 + 30 + 0
= 900230
3. Express the following numbers in standard form:
(i) 5,00,00,000
Solution:-
The standard form of the given number is 5 × 107
(ii) 70,00,000
Solution:-
The standard form of the given number is 7 × 106
(iii) 3,18,65,00,000
Solution:-
The standard form of the given number is 3.1865 × 109
(iv) 3,90,878
Solution:-
The standard form of the given number is 3.90878 × 105
(v) 39087.8
Solution:-
The standard form of the given number is 3.90878 × 104
(vi) 3908.78
Solution:-
The standard form of the given number is 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.
Solution:-
The standard form of the number appearing in the given statement is 3.84 × 108m.
(b) Speed of light in a vacuum is 300,000,000 m/s.
Solution:-
The standard form of the number appearing in the given statement is 3 × 108m/s.
(c) Diameter of the Earth is 1,27,56,000 m.
Solution:-
The standard form of the number appearing in the given statement is 1.2756 × 107m.
(d) Diameter of the Sun is 1,400,000,000 m.
Solution:-
The standard form of the number appearing in the given statement is 1.4 × 109m.
(e) In a galaxy, there are, on average, 100,000,000,000 stars.
Solution:-
The standard form of the number appearing in the given statement is 1 × 1011 stars.
(f) The universe is estimated to be about 12,000,000,000 years old.
Solution:-
The standard form of the number appearing in the given statement is 1.2 × 1010 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.
Solution:-
The standard form of the number appearing in the given statement is 3 × 1020m.
(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.
Solution:-
The standard form of the number appearing in the given statement is 6.023 × 1022 molecules.
(i) The Earth has 1,353,000,000 cubic km of seawater.
Solution:-
The standard form of the number appearing in the given statement is 1.353 × 109 cubic km.
(j) The population of India was about 1,027,000,000 in March 2001.
Solution:-
The standard form of the number appearing in the given statement is 1.027 × 109.
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