Selina Solutions Concise Maths Class 7 Chapter 5: Exponents (Including Laws of Exponents)

Selina Solutions Concise Maths Class 7 Chapter 5 Exponents (Including Laws of Exponents) are designed by a set of highly experienced faculty at BYJU’S. The solutions contain explanations in simple language to improve analytical thinking among students. PDF of solutions are available for free download to boost their exam preparation. Selina Solutions Concise Maths Class 7 Chapter 5 Exponents (Including Laws of Exponents), PDF links are provided here.

Chapter 5 helps students understand the exponential form of a number, laws of exponents and more about indices. Students can refer to the solutions PDF and solve the chapter wise problems on a daily basis, for a better academic score.

Selina Solutions Concise Maths Class 7 Chapter 5: Exponents (Including Laws of Exponents) Download PDF

 

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Exercises of Selina Solutions Concise Maths Class 7 Chapter 5 – Exponents (Including Laws of Exponents)

Exercise 5A Solutions

Exercise 5B Solutions

Access Selina Solutions Concise Maths Class 7 Chapter 5: Exponents (Including Laws of Exponents)

Exercise 5A page: 73

1. Find the value of:

(i) 62

(ii) 73

(iii) 44

(iv) 55

(v) 83

(vi) 75

Solution:

(i) 62

It can be written as

= 6 × 6

= 36

(ii) 73

It can be written as

= 7 × 7 × 7

= 343

(iii) 44

It can be written as

= 4 × 4 × 4 × 4

= 256

(iv) 55

It can be written as

= 5 × 5 × 5 × 5 × 5

= 3125

(v) 83

It can be written as

= 8 × 8 × 8

= 512

(vi) 75

It can be written as

= 7 × 7 × 7 × 7 × 7

= 16807

2. Evaluate:

(i) 23 × 42

(ii) 23 × 52

(iii) 33 × 52

(iv) 22 × 33

(v) 32 × 53

(vi) 53 × 24

(vii) 32 × 42

(viii) (4 × 3)3

(ix) (5 × 4)2

Solution:

(i) 23 × 42

It can be written as

= 2 × 2 × 2 × 4 × 4

On further calculation

= 8 × 16

= 128

(ii) 23 × 52

It can be written as

= 2 × 2 × 2 × 5 × 5

On further calculation

= 8 × 25

= 200

(iii) 33 × 52

It can be written as

= 3 × 3 × 3 × 5 × 5

On further calculation

= 27 × 25

= 675

(iv) 22 × 33

It can be written as

= 2 × 2 × 3 × 3 × 3

On further calculation

= 4 × 27

= 108

(v) 32 × 53

It can be written as

= 3 × 3 × 5 × 5 × 5

On further calculation

= 9 × 125

= 1125

(vi) 53 × 24

It can be written as

= 5 × 5 × 5 × 2 × 2 × 2 × 2

On further calculation

= 125 × 16

= 2000

(vii) 32 × 42

It can be written as

= 3 × 3 × 4 × 4

On further calculation

= 9 × 16

= 144

(viii) (4 × 3)3

It can be written as

= 4 × 4 × 4 × 3 × 3 × 3

On further calculation

= 64 × 27

= 1728

(ix) (5 × 4)2

It can be written as

= 5 × 5 × 4 × 4

On further calculation

= 25 × 16

= 400

3. Evaluate:

(i) (3/4)4

(ii) (-5/6)5

(iii) (-3/-5)3

Solution:

(i) (3/4)4

It can be written as

= (3/4) × (3/4) × (3/4) × (3/4)

On further calculation

= (3 × 3 × 3 × 3)/ (4 × 4 × 4 × 4)

= 81/256

(ii) (-5/6)5

It can be written as

= (-5/6) × (-5/6) × (-5/6) × (-5/6) × (-5/6)

On further calculation

= [(-5) × (-5) × (-5) × (-5) × (-5)]/ (6 × 6 × 6 × 6 × 6)

= -3125/776

(iii) (-3/-5)3

It can be written as

= (-3/-5) × (-3/-5) × (-3/-5)

On further calculation

= [(-3) × (-3) × (-3)]/ [(-5) × (-5) × (-5)]

= 27/125

4. Evaluate:

(i) (2/3)3 × (3/4)2

(ii) (-3/4)3 × (2/3)4

(iii) (3/5)2 × (-2/3)3

Solution:

(i) (2/3)3 × (3/4)2

It can be written as

= (2/3) × (2/3) × (2/3) × (3/4) × (3/4)

On further calculation

= 8/27 × 9/16

= 1/6

(ii) (-3/4)3 × (2/3)4

It can be written as

= (-3/4) × (-3/4) × (-3/4) × (2/3) × (2/3) × (2/3) × (2/3)

On further calculation

= -27/64 × 16/81

= -1/12

(iii) (3/5)2 × (-2/3)3

It can be written as

= (3/5) × (3/5) × (-2/3) × (-2/3) × (-2/3)

On further calculation

= 9/25 × (-8/27)

= -8/75

5. Which is greater:

(i) 23 or 32

(ii) 25 or 52

(iii) 43 or 34

(iv) 54 or 45

Solution:

(i) 23 or 32

It can be written as

23 = 2 × 2 × 2 = 8

32 = 3 × 3 = 9

Hence, 9 is greater than 8 i.e. 32 > 23.

(ii) 25 or 52

It can be written as

25 = 2 × 2 × 2 × 2 × 2 = 32

52 = 5 × 5 = 25

Hence, 32 is greater than 25 i.e. 25 > 52.

(iii) 43 or 34

It can be written as

43 = 4 × 4 × 4 = 64

34 = 3 × 3 × 3 × 3 = 81

Hence, 81 is greater than 64 i.e. 34 > 43.

(iv) 54 or 45

It can be written as

54 = 5 × 5 × 5 × 5 = 625

45 = 4 × 4 × 4 × 4 × 4 = 1024

Hence, 1024 is greater than 625 i.e. 45 > 54.

6. Express each of the following in exponential form:

(i) 512

(ii) 1250

(iii) 1458

(iv) 3600

(v) 1350

(vi) 1176

Solution:

(i) 512

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 1

So we get

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29

(ii) 1250

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 2

So we get

1250 = 2 × 5 × 5 × 5 × 5 = 2 × 54

(iii) 1458

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 3

So we get

1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3 = 2 × 36

(iv) 3600

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 4

So we get

3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24 × 32 × 52

(v) 1350

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 5

So we get

1350 = 2 × 3 × 3 × 3 × 5 × 5 = 2 × 33 × 52

(vi) 1176

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 6

So we get

1176 = 2 × 2 × 2 × 3 × 7 × 7 = 23 × 3 × 72

7. If a = 2 and b = 3, find the value of:

(i) (a + b)2

(ii) (b – a)3

(iii) (a × b)a

(iv) (a × b)b

Solution:

(i) (a + b)2

By substituting the values of a and b

= (2 + 3)2

On further calculation

= 52

= 5 × 5

= 25

(ii) (b – a)3

By substituting the values of a and b

= (3 – 2)3

On further calculation

= 13

= 1 × 1 × 1

= 1

(iii) (a × b)a

By substituting the values of a and b

= (2 × 3)2

On further calculation

= 62

= 6 × 6

= 36

(iv) (a × b)b

By substituting the values of a and b

= (2 × 3)3

On further calculation

= 63

= 6 × 6 × 6

= 216

8. Express:

(i) 1024 as a power of 2.

(ii) 343 as a power of 7.

(iii) 729 as a power of 3.

Solution:

(i) 1024 as a power of 2.

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 7

So we get

1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 210

(ii) 343 as a power of 7.

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 8

So we get

343 = 7 × 7 × 7 = 73

(iii) 729 as a power of 3.

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 9

So we get

729 = 3 × 3 × 3 × 3 × 3 × 3 = 36

9. If 27 × 32 = 3x × 2y; find the values of x and y.

Solution:

It is given that

27 × 32 = 3x × 2y

So we get

27 = 3x

Selina Solutions Concise Maths Class 7 Chapter 5 Image 10

Here

27 = 3 × 3 × 3 = 33 = 3x

We get

x = 3x

Similarly

32 = 2y

Selina Solutions Concise Maths Class 7 Chapter 5 Image 11

Here

32 = 2 × 2 × 2 × 2 × 2 = 25 = 2y

We get

y = 5

10. If 64 × 625 = 2a × 5b; find: (i) the values of a and b. (ii) 2b × 5a.

Solution:

(i) the values of a and b

It is given that

64 × 625 = 2a × 5b

We know that

64 = 2a

Selina Solutions Concise Maths Class 7 Chapter 5 Image 12

We can write it as

64 = 2 × 2 × 2 × 2 × 2 × 2

So we get

64 = 26

a = 6

Similarly

625 = 5b

Selina Solutions Concise Maths Class 7 Chapter 5 Image 13

We can write it as

625 = 5 × 5 × 5 × 5

So we get

625 = 54

b = 4

(ii) 2b × 5a

Substituting the values of a and b

= 24 × 56

It can be written as

= 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5 × 5

So we get

= 16 × 15625

= 250000

Exercise 5B page: 75

1. Fill in the blanks:

(i) In 52 = 25, base = ………. and index = …………

(ii) If index = 3x and base = 2y, the number = ………

Solution:

(i) In 52 = 25, base = 5 and index = 2.

(ii) If index = 3x and base = 2y, the number = 2y3x.

2. Evaluate:

(i) 28 ÷ 23
(ii) 2 28
(iii) (26)0
(iv) (3o)6
(v) 83 x 8-5 x 84
(vi) 5x 53 ÷ 55
(vii) 54 ÷ 53 x 55
(viii) 44 ÷ 43 x 40
(ix) (35 x 47 x 58)0

Solution:

(i) 28 ÷ 23

It can be written as

= 28/ 23

On further calculation

= 2 8-3

= 25

(ii) 2 28

It can be written as

= 23/ 28

On further calculation

= 2 3-8

So we get

= 2 -5

= 1/25

(iii) (26)0

It can be written as

= 2 6 × 0

On further calculation

= 20

So we get

= 1

(iv) (3o) 6

It can be written as

= 3 0 × 6

On further calculation

= 30

So we get

= 1

(v) 83 x 8-5 x 84

It can be written as

= 8 3 + 4 – 5

On further calculation

= 8 7 – 5

So we get

= 82

(vi) 5x 53 ÷ 55

It can be written as

= (5x 53)/ 55

On further calculation

= 5 4 + 3 – 5

So we get

= 5 7 – 5

= 52

(vii) 54 ÷ 53 x 55

It can be written as

= 54/ 53 x 55

On further calculation

= 5 4 – 3 + 5

So we get

= 56

(viii) 44 ÷ 43 x 40

It can be written as

= 44/ (43 x 40)

On further calculation

= 44/ (43 x 1)

So we get

= 44/43

= 44-3

= 41

= 4

(ix) (35 x 47 x 58)0

It can be written as

= 35 × 0 × 47 × 0 × 58 × 0

On further calculation

= 30 × 40 × 50

So we get

= 1 × 1 × 1

= 1

3. Simplify, giving answers with positive index:

(i) 2b6. b3. 5b4

(ii) x2y3. 6x5y. 9x3y4

(iii) (-a)5 (a2)

(iv) (-y)2 (-y)3

(v) (-3)2 (3)3

(vi) (-4x) (-5x2)

(vii) (5a2b) (2ab2) (a3b)

(viii) x2a + 7. x2a – 8

(ix) 3y. 32. 3-4

(x) 24a. 23a. 2-a

(xi) 4x2y2 ÷ 9x3y3

(xii) (102)3 (x8)12

(xiii) (a10)10 (16)10

(xiv) (n2)2 (-n2)3

(xv) – (3ab)2 (-5a2bc4)2

(xvi) (-2)2 × (0)3 × (3)3

(xvii) (2a3)4 (4a2)2

(xviii) (4x2y3)3 ÷ (3x2y3)3

Selina Solutions Concise Maths Class 7 Chapter 5 Image 14

Solution:

(i) 2b6. b3. 5b4

It can be written as

= 2 × 5 × b6 + 3 + 4

On further calculation

= 10 b13

(ii) x2y3. 6x5y. 9x3y4

It can be written as

= 6 × 9 × x2 + 5 + 3 × y3 + 1 + 4

On further calculation

= 54 x10 y8

(iii) (-a)5 (a2)

It can be written as

= (-1 × a)5 × a2

On further calculation

= (-1)5 × a5 + 2

So we get

= – 1 × a7

= – a7

(iv) (-y)2 (-y)3

It can be written as

= (-1 × y)2. (-1 × y)3

On further calculation

= (-1)2. y2. (-1)3 × y3

So we get

= 12 + 3. y2 + 3

= 15 y5

= y5

(v) (-3)2 (3)3

It can be written as

= (-1 × 3)2. (3)3

On further calculation

= (-1)2 × 32. 33

So we get

= (-1)2. 32 + 3

= 1. 35

= 35

(vi) (-4x) (-5x2)

It can be written as

= (-1 × 4 × x). (-1 × 5 × x2)1

On further calculation

= (- 1 × 4 × x). (-1 × 5 × x2)

So we get

= – 1 × – 1 × 4 × 5 × x1 + 2

Here

= – 11 + 1. 41. 51 x3

= 20 x3

(vii) (5a2b) (2ab2) (a3b)

It can be written as

= 5 × 2 × a2 + 1 + 3 × b1 + 2 + 1

On further calculation

= 10 a6b4

(viii) x2a + 7. x2a – 8

It can be written as

= x2a + 7 + 2a – 8

On further calculation

= x4a – 1

(ix) 3y. 32. 3-4

It can be written as

= 3y. 32/34

On further calculation

= 3y. (3 × 3)/ (3 × 3 × 3 × 3)

So we get

= 3y × 1/32

= 3y – 2

(x) 24a. 23a. 2-a

It can be written as

= 2 4a + 3a – a

On further calculation

= 2 7a – a

So we get

= 26a

(xi) 4x2y2 ÷ 9x3y3

It can be written as

= 4x2y2/ 9x3y3

On further calculation

= 4x2 – 3 y2 – 3/ 9

So we get

= 4x-1y-1/ 9

= 4/9xy

(xii) (102)3 (x8)12

It can be written as

= 102 × 3. x8 × 12

On further calculation

= 106 x96

(xiii) (a10)10 (16)10

It can be written as

= a10 × 10. 16 × 10

On further calculation

= a100. 160

So we get

= a100

(xiv) (n2)2 (-n2)3

It can be written as

= n2 × 2. (-n)2 × 3

On further calculation

= n4. (-n)6

So we get

= – n4 – 16 n6

= – n4 + 6

= – n 10

(xv) – (3ab)2 (-5a2bc4)2

It can be written as

= – (32a2b2) × (-1)2 × 52a2 × 2b2c4 × 2

On further calculation

= – (32a2b2) (52a4b2c8)

So we get

= – 32. 52. a2 + 4 b2 + 2 c8

= – 225a6b4c8

(xvi) (-2)2 × (0)3 × (3)3

It can be written as

= 4 × 0 × 27

On further calculation

= 0

(xvii) (2a3)4 (4a2)2

It can be written as

= (2a3)4 (22a2)2

On further calculation

= 24 a3 × 4. 22 × 2. a2 × 2

So we get

= 24 a12. 24 a4

Here

= 24 + 4. a12 + 4

= 28 a16

We get

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × a16

= 256 a16

(xviii) (4x2y3)3 ÷ (3x2y3)3

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 15

Selina Solutions Concise Maths Class 7 Chapter 5 Image 16

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 17

Selina Solutions Concise Maths Class 7 Chapter 5 Image 18

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 19

Selina Solutions Concise Maths Class 7 Chapter 5 Image 20

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 21

Selina Solutions Concise Maths Class 7 Chapter 5 Image 22

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 23

Selina Solutions Concise Maths Class 7 Chapter 5 Image 24

4. Simplify and express the answer in the positive exponent form:

Selina Solutions Concise Maths Class 7 Chapter 5 Image 25

Solution:

Selina Solutions Concise Maths Class 7 Chapter 5 Image 26

= – (3)3 – 1 26 – 4

= – (3)2 22

= – 3222

Selina Solutions Concise Maths Class 7 Chapter 5 Image 27

Selina Solutions Concise Maths Class 7 Chapter 5 Image 28

Selina Solutions Concise Maths Class 7 Chapter 5 Image 29

(iv) – 128/2187

So we get

Selina Solutions Concise Maths Class 7 Chapter 5 Image 30

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 31

Selina Solutions Concise Maths Class 7 Chapter 5 Image 32

(vi) (a3b-5)-2

It can be written as

= a 3 x -2 b-5 x -2

So we get

= a-6 b10

= b10/ a6

5. Evaluate:

(i) 6-2 ÷ (4-2 × 3-2)

Selina Solutions Concise Maths Class 7 Chapter 5 Image 33

(iii) 53 × 32 + (17)0 × 73

(iv) 25 × 150 + (-3)3 – (2/7)-2

(v) (22)0 + 2-4 ÷ 2-6 + (1/2)-3

(vi) 5n × 25n-1 ÷ (5n-1 × 25n-1)

Solution:

(i) 6-2 ÷ (4-2 × 3-2)

It can be written as

= (1/6)2 ÷ (1/4)2 × (1/3)2

On further calculation

= 1/36 ÷ 1/16 × 1/9

So we get

= 1/36 ÷ 1/144

= 1/36 × 144/1

= 4

Selina Solutions Concise Maths Class 7 Chapter 5 Image 34

Selina Solutions Concise Maths Class 7 Chapter 5 Image 35

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 36

Selina Solutions Concise Maths Class 7 Chapter 5 Image 37

(iii) 53 × 32 + (17)0 × 73

It can be written as

= 5 × 5 × 5 × 3 × 3 + (17)0 × 7 × 7 × 7

On further calculation

= 125 × 9 + 1 × 343

So we get

= 1125 + 343

= 1468

(iv) 25 × 150 + (-3)3 – (2/7)-2

It can be written as

= 2 × 2 × 2 × 2 × 2 × 1 + (-3) × (-3) × (-3) – (7/2) × (7/2)

By further calculation

= 32 × 1 – 27 – 49/4

Here we get LCM = 4

= (32 × 4)/ (1 × 4) – (27 × 4)/ (1 × 4) – 49/ (4 × 1)

So we get

= (128 – 108 – 49)/ 4

By subtraction

= -29/ 4

= – 7 1/4

(v) (22)0 + 2-4 ÷ 2-6 + (1/2)-3

It can be written as

= (4)0 + (1/2)4 ÷ (1/2)6 + (2/1)3

By further calculation

= 1 + (1/2 × 1/2 × 1/2 × 1/2) ÷ (1/2 × 1/2 × 1/2 × 1/2 × 1/2 × 1/2) + (2/1 × 2/1 × 2/1)

So we get

= 1 + (1/2 × 1/2 × 1/2 × 1/2 × 2 × 2 × 2 × 2 × 2 × 2) + 8

On further simplification

= 1 + 4 + 8

= 13

(vi) 5n × 25n-1 ÷ (5n-1 × 25n-1)

It can be written as

= 5n × 25n-1 × 1/ (5n-1 × 25n-1)

By further calculation

= 5n × 1/ 5n-1

So we get

= 5n – n + 1

= 51

6. If m = – 2 and n = 2; find the value of:

(i) m2 + n2 – 2mn

(ii) mn + nm

(iii) 6m-3 + 4n2

(iv) 2n3 – 3m

Solution:

(i) m2 + n2 – 2mn

It is given that

m = – 2 and n = 2

Substituting the values we get

= (-2)2 + 22 – 2 (-2) (2)

By further calculation

= 4 + 4 – (-8)

So we get

= 8 + 8

= 16

= 24

Selina Solutions Concise Maths Class 7 Chapter 5 Image 38

(ii) mn + nm

It is given that m = – 2 and n = 2

Substituting the values we get

= (-2)2 + (2)-2

We can write it as

= 4 + 1/2 × 1/2

We get the LCM = 4

= (4 × 4)/ (1 × 4) + 1/4

So we get

= (16 + 1)/ 4

= 17/ 4

= 4 ¼

(iii) 6m-3 + 4n2

It is given that

m = – 2 and n = 2

Substituting the values

= 6 (-2)-3 + 4 (2)2

It can be written as

= 6 × 1/ -2 × 1/ -2 × 1/ -2 + 4 × 2 × 2

So we get

= – 3/4 + 16

Here the LCM = 4

= (-3 + 16 × 4)/ 4

By calculation

= (-3 + 64)/ 4

= – 61/4

= 15 ¼

(iv) 2n3 – 3m

It is given that

m = – 2 and n = 2

By substituting the values

= 2 (2)3 – 3 (-2)

It can be written as

= 2 × (2 × 2 × 2) – 3 × (-2)

By further calculation

= 16 – 3 × (-2)

So we get

= 16 + 6

= 22

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