 # NCERT Solutions for Class 7 Maths Exercise 13.1 Chapter 13 Exponents and Powers

NCERT Solutions for Class 7 Maths Exercise 13.1 Chapter 13 Exponents and Powers in simple PDF are available here. This exercise of NCERT Solutions for Class 7 Maths Chapter 13 contains topics related to exponents. This will help students to have a better understanding of the concepts and they are able to develop problem-solving abilities. This NCERT Solutions for Class 7 Chapter 13, can also be used by the students who appear for any competitive exams.

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Exercise 13.2 Solutions

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### Access Answers to NCERT Class 7 Maths Chapter 13 – Exponents and Powers Exercise 13.1

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 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 =

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 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

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1. Yasir Bashir