# NCERT solution class 7 chapter 13 Exponents and powers 13.1 mathematics

EXERCISE 13.1

#### Question 1:

Find the value of:

(i) 26 (ii) 93

(iii) 112 (iv)54

(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 × d

(i) 6 × 6 × 6 × 6 = 64

(ii) t × tt2

(iii) b × × × b4

(iv) 5 × 5 × 7 × 7 × 7 = 52 × 73

(v) 2 × 2 × a × = 22 × a2

(vi) a × a × a × × c × c × c × a3 c4 d

#### Question 3:

Express the following numbers using exponential notation:

(i) 512 (ii) 343

(iii) 729 (iv) 3125

(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) 100or 2100

(v) 210 or 102

(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

(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 × 10­­­­­ (vi) 52 × 33

(vii) 2× 32 (viii) 32 × 104

(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

(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

(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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