Class – 7 CH-13 EXPONENTS AND POWERS

MATHS NCERT SOLUTIONS

Exercise 13.1

Question 1:

Find the value of:

(i) 26 (ii) 93 (iii) 112 (iv) 54

SOLUTION 1:

(i) 26 = 2 x 2 x 2 x 2 x 2 x 2 = 64

(ii) 93 = 9 x 9 x 9 = 729

(iii) 112 = 11 x 11 = 121

(iv) 54= 5 x 5 x 5 x 5 = 625

Question 2:

Express the following in exponential form:

(i) 6 x 6 x 6 x 6 (ii) tt

(iii) b b b b   (iv) 5 x 5 x 7 x 7 x 7

(v) 2 2  a a (vi) a a a c c c c d      

SOLUTION 2:

(i) 6 x 6 x 6 x 6 = 64

(ii) t t  t2

(iii) b b b b    b4

(iv) 5 x 5 x 7 x 7 x 7 = 52 x 73

(v) 2    2 a a 22 a2

(vi) a a a c c c c d         a3 c4 d

Question 3:

Express each of the following numbers using exponential notation:

(i) 512 (ii) 343 (iii) 729 (iv) 3125

SOLUTION 3:

(i) 512

(ii)

(iii)

(iv)

2 512

2 256

2 128

2 64

2 32

2 16

2 8

2 4

2 2

512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 29

343

7 343

7 49

7 7

343 = 7 x 7 x 7 = 73

729

3 729

3 243

3 81

3 27

3 9

3 3

729 = 3 x 3 x 3 x 3 x 3 x 3 = 36

3125

5 3125

5 625

5 125

5 25

5 5

3125 = 5 x 5 x 5 x 5 x 5

Question 4:

Identify the greater number, wherever possible, in each of the following:

(i) 43 and 34 (ii) 53 or 35

(iii) 28 or 82 (iv) 1002 or 2100

(v) 210 or 102 SOLUTION 4:

(i) 43 = 4 x 4 x 4 = 64

34 = 3 x 3 x 3 x 3 = 81

Since 64 < 81

Thus, 34 is greater than 43.

(ii) 53 = 5 x 5 x 5 = 125

35 = 3 x 3 x 3 x 3 x 3 = 243

Since, 125 < 243

Thus, 34 is greater than 53.

(iii) 28 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256

82 = 8 x 8 = 64

Since, 256 > 64

Thus, 28 is greater than 82.

(iv) 1002 = 100 x 100 = 10,000

2100 = 2 x 2 x 2 x 2 x 2 x …..14 times x ……… x 2 = 16,384 x ….. x 2

Since, 10,000 < 16,384 x ……. x 2 Thus, 2100 is greater than 1002.

(v) 210 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1,024

102 = 10 x 10 = 100

Since, 1,024 > 100

Thus, 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

SOLUTION 5:

(i) 648 = 23 x 34

2 648

2 324

2 162

3 81

3 27

3 9

3 3

(ii) 405 = 5 x 34

5 405

3 81

3 27

3 9

3 3

(iii) 540 = 22 x 33 x 5

2 540

2 270

3 135

3 45

3 15

5 5

(iv) 3,600 = 24 x 32 x 52

Question 6:

Simplify:

(i) 2 x 103 (ii)

(iii) 23 x 5 (iv)

(v) 0 x 102 (vi) (vii) 24 x 32 (viii)

SOLUTION 6:

(i) 2 x 103 = 2 x 10 x 10 x 10

(ii) 72 x 22 = 7 x 7 x 2 x 2

(iii) 23 x 5 = 2 x 2 x 2 x 5

(iv) 3 x 44 = 3 x 4 x 4 x 4 x 4

(v) 0 x 102 = 0 x 10 x 10

(vi) 53 x 33 = 5 x 5 x 3 x 3 x 3

(vii) 24 x 32 = 2 x 2 x 2 x 2 x 3 x 3

(viii) 32 x 104 = 3 x 3 x 10 x 10 x 10 x 10

Question 7:

Simplify:

(i) 43 (ii)

(iii) 32  52 (iv)

SOLUTION 7:

(i) 43        4  4  4 64

2 3600

2 1800

2 900

2 450

3 225

3 75

5 25

5 5

72 x 22 3 x 44

52 x 33

32 x 104

= 2,000

= 196

= 40

= 768

= 0

= 675

= 144

= 90,000

  3  23

  23  103

(ii)   3  23          3  2  2  2 24

(iii) 32  52          3  3  5  5 225

(iv) 23  103         2  2  2  10  10  10

Question 8:

Compare the following numbers:

(i) 2.7 x 1012; 1.5 x 108 (ii) 4 x 1014; 3 x 1017 SOLUTION 8:

(i) 2.7 x 1012 and 1.5 x 108

On comparing the exponents of base 10,

2.7 x 1012 > 1.5 x 108

(ii) 4 x 1014 and 3 x 1017

On comparing the exponents of base 10,

4 x 1014 < 3 x 1017

Exercise 13.2

Question 1:

Using laws of exponents, simplify and write the SOLUTION in exponential form:

(i) 32 x 34 x 38 (ii) 615  610

(iii) a3a2 (iv) 7x 72

(v) (52)2  53 (vi) 25 x 55

(vii) a b4  4 (viii) (34)3

(ix) (220  215) x 23 (x) 8 8t  2

SOLUTION 1:

(i) 3 3 3 32  4 8 2 4 8   314  a a am n  m n 

(ii) 615  610 615 10  65  a a am  n  m n 

(iii) a a a3 2 3 2 a5  a a am n  m n 

(iv) 7x  72 7x2  a a am n  m n 

(v) 523  5 53 2 3   5 5 53 6 3  amn am n 

= 56 3 53  a a am  n  m n 

(vi) 2555 2 5 5 105  a bm m a b m

(vii) a b4 4 a b 4  a bm m a b m

(viii) 343 = 34 3  312  amn am n 

(ix) 220 21523 = 220 15 23  a a am  n  m n 

= 2 2 25 3 5 3 28  a a am n  m n 

(x) 8t  82 8t2  a a am  n  m n 

Question 2:

Simplify and express each of the following in exponential form:

(i) 2 3 433 32  4 (ii) 5235457



4 53 (iv) 3 7 11 2 8

(iii) 25 

21 11

(v) 37 0  30 40

(vi) 2

3 34  3

(vii) 20  30 40 (viii) 30 2050

28a5 a5  8

(ix) 43a3 (x)  a3 a

(xi) 4455a ba b5 28 3 (xii)  3 22 2 

SOLUTION 2:

2 3 4 2 3 23 4 3 4 2 23 2 34

(i) 3 32  3 2 5  3 2 5  a a am n  m n 

3 2

= 20 33 = 1 3 3 33

3 n

=  54 25 5 34 3  a a am  n  m n  2 35

(ii) 52 5457 565457  am am n 

= 56 4 57  510 57  a a am n  m n 

= 510 7  53  a a am  n  m n 

(iii) 25 54  3 524   5 5 53 8 3  amn am n 

= 58 3 55  a a am  n  m n 

3 7 11 3 7 11

(iv)  21 112 3 8  3 7 11  2 38  31 1 72 1 118 3  a a am  n  m n 

= 30  71 115 = 7 11 5

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

37 37 37

  3 3 34  3 4 3 37

= 37 7  30 1

20      30 40 1 1 1 3

20      30 40 1 1 1 1

30 2050      1 1 1 2 1 2

28a5 28a5 28a5

3 a3  223a3 26a3 4 

= 28 6 a5 2  22 a2

= 2a2

 a5  8 a5 3 a8  a2 a8

 a3 a 

= a2 8  a10

45a b8 3 45 5 a8 5 b3 2   40 a3 b

 45a b5 2

= 1  a b ab3 3

2 23 2 23 1 2 242

= 24 2 28















 a a am n  m n  a a am  n  m n 

a0 1

amn am n 

a a am  n  m n  a bm m a b m

a a am  n  m n  a a am n  m n 

a a am  n  m n  a0 1

a a am n  m n 

Question 3:

Say true or false and justify your SOLUTION:

(i) 10 x 1011 = 10011 (ii) 23 > 52

(iii) 23 x 32 = 65 (iv) 30 = (1000)0

SOLUTION 3:

(i) 10 10 11 10011

L.H.S. 101 11 = 1012 and R.H.S. 10211 1022

Since, L.H.S.  R.H.S.

Therefore, it is false.

(ii) 23  52

L.H.S. 2 83  and R.H.S. 52  25

Since, L.H.S. is not greater than R.H.S.

Therefore, it is false.

(iii) 23 32 65

L.H.S. 23   32 8 9 72 and R.H.S. 65  7,776

Since, L.H.S.  R.H.S.

Therefore, it is false.

(iv) 30 10000

L.H.S. 30 1 and R.H.S. 10000 = 1

Since, L.H.S. = R.H.S.

Therefore, it is true.

Question 4:

Express each of the following as a product of prime factors only in exponential form:

(i) 108 x 192 (ii) 270 (iii) 729 x 64 (iv) 768

SOLUTION 4:

(i) 108 x 192

(ii)

(iii)

108 x 192

270

729 x 64

729 x 64 2 192

2 96

2 48

2 24

2 12

2 6

3 3

= 2233263

= 22 6 33 1

= 28 34

2 108

2 54

3 27

3 9

3 3

2 270

3 135

3 45

3 15

5 5

= 2 35 5

= 36 26

2 64

2 32

2 16

2 8

2 4

2 2

3 729

3 243

3 81

3 27

3 9

3 3

(iv) 768

2 768

2 384

2 192

2 96

2 48

2 24

2 12

2 6

3 3

768 = 28 3

Question 5:

Simplify:

252 73

(i) 3

8 7

25 5

(ii)  3 2t4t8

10 

3 10 255

(iii)  7 5 5

5 6

SOLUTION 5:

25273 25 2 73

(i) 837  2337

2 710  3

= 2 79 

= 210 9 73 1  2 72

= 2 x 49

= 98

25 5 2 t8 5 52 2 t8

(ii) 103t4  5 2 3t4

52 2 t8 4

= 2 33 3

54 t4

= 2 53 3

54 3 t4

= 23

5t4

3 10 255 5 3 2 5 55   5 2

(iii) 5 67  5 = 57  2 35

3 2 5 55  5 5 2

= 5 2 37  5 5

3 2 55 5 5 2

= 5 2 37  5 5

3 2 55 5 7

= 5 2 37  5 5

= 25 5 35 5 55 5

= 20  30 50

= 1 x 1 x 1

= 1

Exercise 13.3

Question 1:

Write the following numbers in the expanded form:

279404, 3006194, 2806196, 120719, 20068

SOLUTION 1:

(i) 2,79,404 = 2,00,000 + 70,000 + 9,000 + 400 + 00 + 4

= 2 x 100000 + 7 x 10000 + 9 x 1000 + 4 x 100 + 0 x 10 + 4 x 1

= 2 10          5 7 104 9 103 4 102 0 101 4 100 (ii) 30,06,194 = 30,00,000 + 0 + 0 + 6,000 + 100 + 90 + 4

= 3 x 1000000 + 0 x 100000 + 0 x 10000 + 6 x 1000 + 1 x 100 + 9 x 10 + 4 x 1

= 3 10            6 0 105 0 104 6 103 1 102 9 10 4 100

(iii) 28,06,196 = 20,00,000 + 8,00,000 + 0 + 6,000 + 100 + 90 + 6

= 2 x 1000000 + 8 x 100000 + 0 x 10000 + 6 x 1000 + 1 x 100 + 9 x 10 + 6 x 1

= 2 10            6 8 105 0 104 6 103 1 102 9 10 6 100

(iv) 1,20,719 = 1,00,000 + 20,000 + 0 + 700 + 10 + 9

= 1 x 100000 + 2 x 10000 + 0 x 1000 + 7 x 100 + 1 x 10 + 9 x 1

= 1 10          5 2 104 0 103 7 102 1 101 9 100

(v) 20,068 = 20,000 + 00 + 00 + 60 + 8

= 2 x 10000 + 0 x 1000 + 0 x 100 + 6 x 10 + 8 x 1

= 2 10        4 0 103 0 102 6 101 8 100

Question 2:

Find the number from each of the following expanded forms:

(a) 8 x 104 + 6 x 103 + 0 x 102 + 4 x 101 + 5 x 100

(b) 4 x 105 + 5 x 103 + 3 x 102 + 2 x 100

(d) 9 x 105 + 2 x 102 + 3 x 101 SOLUTION 2:

(a) 8 x 104 + 6 x 103 + 0 x 102 + 4 x 101 + 5 x 100

= 8 x 10000 + 6 x 1000 + 0 x 100 + 4 x 10 + 5 x 1

= 80000 + 6000 + 0 + 40 + 5

= 86,045

(b) 4 x 105 + 5 x 103 + 3 x 102 + 2 x 100

= 4 x 100000 + 0 x 10000 + 5 x 1000 + 3 x 100 + 0 x 10 + 2 x 1

= 400000 + 0 + 5000 + 3000 + 0 + 2

= 4,05,302

= 3 x 10000 + 0 x 1000 + 7 x 100 + 0 x 10 + 5 x 1

= 30000 + 0 + 700 + 0 + 5

= 30,705

(d) 9 x 105 + 2 x 102 + 3 x 101

= 9 x 100000 + 0 x 10000 + 0 x 1000 + 2 x 100 + 3 x 10 + 0 x 1

= 900000 + 0 + 0 + 200 + 30 + 0

= 9,00,230

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

SOLUTION 3:

(i) 5,00,00,000 = 5 x 1,00,00,000 = 5 10 7

(ii) 70,00,000 = 7 x 10,00,000 = 7 10 6

(iii) 3,18,65,00,000 = 31865 x 100000

= 3.1865 x 10000 x 100000 = 3.1865 10 9

(iv) 3,90,878 = 3.90878 x 100000 = 3.90878 10 5

(v) 39087.8 = 3.90878 x 10000 = 3.90878 10 4

(vi) 3908.78 = 3.90878 x 1000 = 3.90878 10 3

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.

(d) Diameter of the Sun is 1,400,000,000 m.

(e) In a galaxy there are on an average 100,000,000,0000 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.

SOLUTION 4:

(a) The distance between Earth and Moon = 384,000,000 m

= 384 x 1000000 m

= 3.84 x 100 x 1000000

= 3.84 10 8 m

(b) Speed of light in vacuum = 300,000,000 m/s

= 3 x 100000000 m/s

= 3 10 8 m/s

= 12756 x 1000 m

= 1.2756 x 10000 x 1000 m

= 1.2756 10 7 m

(d) Diameter of the Sun = 1,400,000,000 m

= 14 x 100,000,000 m

= 1.4 x 10 x 100,000,000 m

= 1.4 10 9 m

(e) Average of Stars = 100,000,000,000

= 1 x 100,000,000,000

= 1 10 11

(f) Years of Universe = 12,000,000,000 years

= 12 x 1000,000,000 years

= 1.2 x 10 x 1000,000,000 years

= 1.2 10 10 years

(g) Distance of the Sun from the centre of the Milky Way Galaxy

= 300,000,000,000,000,000,000 m

= 3 x 100,000,000,000,000,000,000 m

= 3 10 20 m

(h) Number of molecules in a drop of water weighing 1.8 gm

= 60,230,000,000,000,000,000,000

= 6023 x 10,000,000,000,000,000,000

= 6.023 x 1000 x 10,000,000,000,000,000,000

= 6.023 10 22

(i) The Earth has Sea water

(j) The population of India

= 1,353,000,000 km3

= 1,353 x 1000000 km3

= 1.353 x 1000 x 1000,000 km3 = 1.353 10 9 km3

= 1,027,000,000

= 1027 x 1000000

= 1.027 x 1000 x 1000000

= 1.027 10 9

🌈⚡Key To Enjoy Learning Maths☘🌈

Pages