NCERT Solutions for Class 7 Maths Chapter 13 - Exponents and Powers
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Students can access the NCERT Solutions for Class 7 Maths Chapter 13 - Exponents and Powers. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 13.1
Express each of the following numbers using the exponential notation.
(i) 512
(ii) 343
(iii) 729
(iv) 3125
(i) The factors of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
So, it can be expressed in the exponential form of 29.
(ii) The factors of 343 = 7 × 7 × 7
So, it can be expressed in the exponential form of 73.
(iii) The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3
So, it can be expressed in the exponential form of 36.
(iv) The factors of 3125 = 5 × 5 × 5 × 5 × 5
So, it can be expressed in the exponential form of 55.
Express each of the following as a product of powers of their prime factors.
(i) 648
(ii) 405
(iii) 540
(iv) 3,600
(i) Factors of 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3
= 23 × 34
(ii) Factors of 405 = 3 × 3 × 3 × 3 × 5
= 34 × 5
(iii) Factors of 540 = 2 × 2 × 3 × 3 × 3 × 5
= 22 × 33 × 5
(iv) Factors of 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 24 × 32 × 52
Simplify:
(i) (– 4)3
(ii) (–3) × (–2)3
(iii) (–3)2 × (–5)2
(iv) (–2)3 × (–10)3
(i) The expansion of -43
= – 4 × – 4 × – 4
= – 64
(ii) The expansion of (-3) × (-2)3
= – 3 × – 2 × – 2 × – 2
= – 3 × – 8
= 24
(iii) The expansion of (-3)2 × (-5)2
= – 3 × – 3 × – 5 × – 5
= 9 × 25
= 225
(iv) The expansion of (-2)3 × (-10)3
= – 2 × – 2 × – 2 × – 10 × – 10 × – 10
= – 8 × – 1000
= 8000
Compare the following numbers.
(i) 2.7 × 1012 ; 1.5 × 108
(ii) 4 × 1014 ; 3 × 1017
(i) By observing the question,
Comparing the exponents of base 10,
Clearly,
2.7 × 1012 > 1.5 × 108
(ii) By observing the question
Comparing the exponents of base 10,
Clearly,
4 × 1014 < 3 × 1017
Find the value of:
(i) 26
(ii) 93
(iii) 112
(iv) 54
(i) The above value can be written as,
= 2 × 2 × 2 × 2 × 2 × 2
= 64
(ii) The above value can be written as,
= 9 × 9 × 9
= 729
(iii) The above value can be written as,
= 11 × 11
= 121
(iv) The above value can be written as,
= 5 × 5 × 5 × 5
= 625
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
(i) The given question can be expressed in the exponential form as 64.
(ii) The given question can be expressed in the exponential form as t2.
(iii) The given question can be expressed in the exponential form as b4.
(iv) The given question can be expressed in the exponential form as 52 × 73.
(v) The given question can be expressed in the exponential form as 22 × a2.
(vi) The given question can be expressed in the exponential form as a3 × c4 × d.
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
(i) 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) 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) 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) 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 =
Clearly,
1002 < 2100
Hence, 2100 is the greater number.
(v) 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.
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
(i) The above question can be written as,
= 2 × 10 × 10 × 10
= 2 × 1000
= 2000
(ii) The above question can be written as,
= 7 × 7 × 2 × 2
= 49 × 4
= 196
(iii) The above question can be written as,
= 2 × 2 × 2 × 5
= 8 × 5
= 40
(iv) The above question can be written as,
= 3 × 4 × 4 × 4 × 4
= 3 × 256
= 768
(v) The above question can be written as,
= 0 × 10 × 10
= 0 × 100
= 0
(vi) The above question can be written as,
= 5 × 5 × 3 × 3 × 3
= 25 × 27
= 675
(vii) The above question can be written as,
= 2 × 2 × 2 × 2 × 3 × 3
= 16 × 9
= 144
(viii) The above question can be written as,
= 3 × 3 × 10 × 10 × 10 × 10
= 9 × 10000
= 90000
Exercise 13.2
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) (52)3 ÷ 53
(vi) 25 × 55
(vii) a4 × b4
(viii) (34)3
(ix) (220 ÷ 215) × 23
(x) 8t ÷ 82
(i) By the rule of multiplying the powers with the same base = am × an = am + n
Then,
= (3)2 + 4 + 8
= 314
(ii) By the rule of dividing the powers with the same base = am ÷ an = am – n
Then,
= (6)15 – 10
= 65
(iii) By the rule of multiplying the powers with the same base = am × an = am + n
Then,
= (a)3 + 2
= a5
(iv) By the rule of multiplying the powers with the same base = am × an = am + n
Then,
= (7)x + 2
(v) 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 same base = am ÷ an = am – n
Then,
= (5)6 – 3
= 53
(vi) By the rule of multiplying the powers with the same exponents = am × bm = abm
Then,
= (2 × 5)5
= 105
(vii) By the rule of multiplying the powers with the same exponents = am × bm = abm
Then,
= (a × b)4
= ab4
(viii) By the rule of taking the power of a power = (am)n = amn
(34)3 can be written as = (3)4 × 3
= 312
(ix) 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) By the rule of dividing the powers with the same base = am ÷ an = am – n
Then,
= (8)t – 2
Say true or false and justify your answer:
(i) 10 × 1011 = 10011
(ii) 23 > 52
(iii) 23 × 32 = 65
(iv) 30 = (1000)0
(i) 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) 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) 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) 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.
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
(i) 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) The factors of 270 = 2 × 3 × 3 × 3 × 5
= 2 × 33 × 5
(iii) 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) The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
= 28 × 3
Simplify:
(i) ((25)2 × 73)/ (83 × 7)
(ii) (25 × 52 × t8)/ (103 × t4)
(iii) (35 × 105 × 25)/ (57 × 65)
(i) 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 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) 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
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 × 113)
(v) 37/ (34 × 33)
(vi) 20 + 30 + 40
(vii) 20 × 30 × 40
(viii) (30 + 20) × 50
(ix) (28 × a5)/ (43 × a3)
(x) (a5/a3) × a8
(xi) (45 × a8b3)/ (45 × a5b2)
(xii) (23 × 2)2
(i) 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 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) (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) 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+3) … [∵am × an = am + n]
= 37/ 37
= 37 – 7 … [∵am ÷ an = am – n]
= 30
= 1
(vi) = 1 + 1 + 1
= 3
(vii) = 1 × 1 × 1
= 1
(viii) = (1 + 1) × 1
= (2) × 1
= 2
(ix) (4)3 can be written as = (2 × 2)3
= (22)3
(52)4 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
= 2a2 … [∵(am)n = amn]
(x) = (a5 – 3) × a8 … [∵am ÷ an = am – n]
= a2 × a8
= a2 + 8 … [∵am × an = am + n]
= a10
(xi) = 45 – 5 × (a8 – 5 × b3 – 2) … [∵am ÷ an = am – n]
= 40 × (a3b)
= 1 × a3b
= a3b
(xii) = (23 + 1)2 … [∵am × an = am + n]
= (24)2
(24)2 can be written as = (2)4 × 2 … [∵(am)n = amn]
= 28
Exercise 13.3
Write the following numbers in the expanded forms:
(a) 279404
(b) 3006194
(c) 2806196
(d) 120719
(e) 20068
(a) The expanded form of the number 279404
= (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) The expanded form of the number 3006194
= (3 × 1000000) + (0 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + 4
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) The expanded form of the number 2806196
= (2 × 1000000) + (8 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + 6
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) The expanded form of the number 120719
= (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) The expanded form of the number 20068
= (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).
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 a 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 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 seawater.
(j) The population of India was about 1,027,000,000 in March 2001.
(a) The standard form of the number appearing in the given statement is 3.84 × 108m.
(b) The standard form of the number appearing in the given statement is 3 × 108m/s.
(c) The standard form of the number appearing in the given statement is 1.2756 × 107m.
(d) The standard form of the number appearing in the given statement is 1.4 × 109m.
(e) The standard form of the number appearing in the given statement is 1 × 1011.
(f) The standard form of the number appearing in the given statement is 1.2 × 1010.
(g) The standard form of the number appearing in the given statement is 3 × 1020m.
(h) The standard form of the number appearing in the given statement is 6.023 × 1022.
(i) The standard form of the number appearing in the given statement is 1.353 × 109 cubic km.
(j) The standard form of the number appearing in the given statement is 1.027 × 109.
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
(i) The standard form of the given number is 5 × 107
(ii) The standard form of the given number is 7 × 106
(iii) The standard form of the given number is 3.1865 × 109
(iv) The standard form of the given number is 3.90878 × 105
(v) The standard form of the given number is 3.90878 × 104
(vi) The standard form of the given number is 3.90878 × 103
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
b) (4 ×10)5 + (5 × 10)3 + (3 × 10)2 + (2 × 10)0
(c) (3 ×10)4 + (7 × 10)2 + (5 × 10)0
(d) (9 ×10)5 + (2 × 10)2 + (3 × 10)1
(a) The expanded form
= (8 × 10000) + (6 × 1000) + (0 × 100) + (4 × 10) + (5 × 1)
= 80000 + 6000 + 0 + 40 + 5
= 86045
(b) The expanded form
= (4 × 100000) + (0 × 10000) + (5 × 1000) + (3 × 100) + (0 × 10) + (2 × 1)
= 400000 + 0 + 5000 + 300 + 0 + 2
= 405302
(c) The expanded form
= (3 × 10000) + (0 × 1000) + (7 × 100) + (0 × 10) + (5 × 1)
= 30000 + 0 + 700 + 0 + 5
= 30705
(d) The expanded form is,
= (9 × 100000) + (0 × 10000) + (0 × 1000) + (2 × 100) + (3 × 10) + (0 × 1)
= 900000 + 0 + 0 + 200 + 30 + 0
= 900230
Frequently Asked Questions
The NCERT solution for Class 7 Chapter 13: Exponents and Powers is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 7 Chapter 13: Exponents and Powers is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Exponents and Powersally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 7 Maths Chapter 13 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 7 Maths Chapter 13: Exponents and Powers as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 7 Maths Chapter 13 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.