Mathematics NCERT Grade 8, Chapter 13: Exponents and Powers- This chapter highlights all the concepts related to exponents and powers. Initially, the concept of exponents will be discussed where a student will study how to read, write and represent exponents and powers.
Students will know about the importance and use of exponents.
  • The short notation 104 stands for the product 10×10×10×10. Here ‘10’ is called the base and ‘4’ the exponent. The number 104 is read as 10 raised to the power of 4 or simply as the fourth power of 10. 104 is called the exponential form of 10,000.
The other portion of the chapter gives insight about Laws of exponents. This particular section is divided into several sub-sections:
  • Multiplying Powers with the Same Base
  • Dividing Powers with the Same Base
  • Taking Power of a Power
  • Multiplying Powers with the Same Exponents
  • Dividing Powers with the Same Exponents
  • Numbers with exponent zero
Once the student has come across various laws of exponents miscellaneous examples using the laws of exponents can be studied.
This is followed by Decimal Number System and expressing large numbers in the standard form.
  • Any number can be expressed as a decimal number between 1.0 and 10.0 including 1.0 multiplied by a power of 10. Such a form of a number is called its standard form
The chapter includes 3 unsolved exercises and various solved examples.
The chapter comprises a summary in which all the important concepts of the chapter- Exponents and Powers are mentioned.

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

Page No 252:

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

Page No 253:

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

Page No 253:

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
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Video Solution for integers (Page: 253 , Q.No.: 5)

NCERT Solution for Class 7 math - integers 253 , Question 5

Page No 253:

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

Page No 253:

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

Page No 253:

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
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Video Solution for integers (Page: 253 , Q.No.: 8)

NCERT Solution for Class 7 math - integers 253 , Question 8



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

(vii) a4 × b4 (viii) (34)3

(ix) (x) 8t ÷ 82

Answer:

(i) 32 × 34 × 38 = (3)2 + 4 + 8 (am × an = am+n)

= 314

(ii) 615 ÷ 610 = (6)15 − 10 (am ÷ an = amn)

= 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 = amn)

= 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 = amn)

= 25 × 23

= (25 + 3) (am × an = am+n)

= 28

(x) 8t ÷ 82 = 8(t − 2) (am ÷ an = amn)

Page No 260:

Question 2:

Simplify and express each of the following in exponential form:

(i) (ii) (iii)

(iv) (v) (vi) 20 + 30 + 40

(vii) 20 × 30 × 40 (viii) (30 + 20) × 50 (ix)

(x) (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 = amn)

= 53

(iii) 254 ÷ 53 = (5 ×5)4 ÷ 53

= (52)4 ÷ 53

= 52 × 4 ÷ 53 (am)n = amn

= 58 ÷ 53

= 58 − 3 (am ÷ an = amn)

= 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)

= (24)2 = 24 × 2 (am)n = amn

= 28

Page No 260:

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
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(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

Video Solution for integers (Page: 261 , Q.No.: 4)

NCERT Solution for Class 7 math - integers 261 , Question 4

Page No 261:

Question 5:

Simplify:

(i) (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

Page No 263:

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

Page No 263:

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

Page No 263:

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



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