NCERT solutions for class 7 maths chapter 13Exponents and Powers

(i) The factors of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

So, it can be expressed in the exponential form of 29.

(ii) The factors of 343 = 7 × 7 × 7

So, it can be expressed in the exponential form of 73.

(iii) The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

So, it can be expressed in the exponential form of 36.

(iv) The factors of 3125 = 5 × 5 × 5 × 5 × 5

So, it can be expressed in the exponential form of 55.

 

(i) Factors of 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3

= 23 × 34

(ii) Factors of 405 = 3 × 3 × 3 × 3 × 5

= 34 × 5

(iii) Factors of 540 = 2 × 2 × 3 × 3 × 3 × 5

= 22 × 33 × 5

(iv) Factors of 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= 24 × 32 × 52

 

(i) The expansion of -43

= – 4 × – 4 × – 4

= – 64

(ii) The expansion of (-3) × (-2)3

= – 3 × – 2 × – 2 × – 2

= – 3 × – 8

= 24

(iii) The expansion of (-3)2 × (-5)2

= – 3 × – 3 × – 5 × – 5

= 9 × 25

= 225

(iv) The expansion of (-2)3 × (-10)3

= – 2 × – 2 × – 2 × – 10 × – 10 × – 10

= – 8 × – 1000

= 8000

 

(i) By observing the question,

Comparing the exponents of base 10,

Clearly,

2.7 × 1012 > 1.5 × 108

(ii) By observing the question

Comparing the exponents of base 10,

Clearly,

4 × 1014 < 3 × 1017

 

(i) The above value can be written as,

= 2 × 2 × 2 × 2 × 2 × 2

= 64

(ii) The above value can be written as,

= 9 × 9 × 9

= 729

(iii) The above value can be written as,

= 11 × 11

= 121

(iv) The above value can be written as,

= 5 × 5 × 5 × 5

= 625

 

(i) The given question can be expressed in the exponential form as 64.

(ii) The given question can be expressed in the exponential form as t2.

(iii) The given question can be expressed in the exponential form as b4.

(iv) The given question can be expressed in the exponential form as 52 × 73.

(v) The given question can be expressed in the exponential form as 22 × a2.

(vi) The given question can be expressed in the exponential form as a3 × c4 × d.

 

(i) 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) 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) 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) 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) 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.

 

(i) The above question can be written as,

= 2 × 10 × 10 × 10

= 2 × 1000

= 2000

(ii) The above question can be written as,

= 7 × 7 × 2 × 2

= 49 × 4

= 196

(iii) The above question can be written as,

= 2 × 2 × 2 × 5

= 8 × 5

= 40

(iv) The above question can be written as,

= 3 × 4 × 4 × 4 × 4

= 3 × 256

= 768

(v) The above question can be written as,

= 0 × 10 × 10

= 0 × 100

= 0

(vi) The above question can be written as,

= 5 × 5 × 3 × 3 × 3

= 25 × 27

= 675

(vii) The above question can be written as,

= 2 × 2 × 2 × 2 × 3 × 3

= 16 × 9

= 144

(viii) The above question can be written as,

= 3 × 3 × 10 × 10 × 10 × 10

= 9 × 10000

= 90000

 

(i) By the rule of multiplying the powers with the same base = am × an = am + n

Then,

= (3)2 + 4 + 8

= 314

(ii) By the rule of dividing the powers with the same base = am ÷ an = am – n

Then,

= (6)15 – 10

= 65

(iii) By the rule of multiplying the powers with the same base = am × an = am + n

Then,

= (a)3 + 2

= a5

(iv) By the rule of multiplying the powers with the same base = am × an = am + n

Then,

= (7)x + 2

(v) By the rule of taking the power of as power = (am)n = amn

(52)3 can be written as = (5)2 × 3

= 56

Now, 56 ÷ 53

By the rule of dividing the powers with same base = am ÷ an = am – n

Then,

= (5)6 – 3

= 53

(vi) By the rule of multiplying the powers with the same exponents = am × bm = abm

Then,

= (2 × 5)5

= 105

(vii) By the rule of multiplying the powers with the same exponents = am × bm = abm

Then,

= (a × b)4

= ab4

(viii) By the rule of taking the power of a power = (am)n = amn

(34)3 can be written as = (3)4 × 3

= 312

(ix) By the rule of dividing the powers with the same base = am ÷ an = am – n

(220 ÷ 215) can be simplified as,

= (2)20 – 15

= 25

Then,

By the rule of multiplying the powers with the same base = am × an = am + n

25 × 23 can be simplified as,

= (2)5 + 3

= 28

(x) By the rule of dividing the powers with the same base = am ÷ an = am – n

Then,

= (8)t – 2

 

(i) Let us consider Left Hand Side (LHS) = 10 × 1011

= 101 + 11 … [∵am × an = am + n]

= 1012

Now, consider Right Hand Side (RHS) = 10011

= (10 × 10)11

= (101 + 1)11

= (102)11

= (10)2 × 11 … [∵(am)n = amn]

= 1022

By comparing LHS and RHS,

LHS ≠ RHS

Hence, the given statement is false.

(ii) Let us consider LHS = 23

Expansion of 23 = 2 × 2 × 2

= 8

Now, consider RHS = 52

Expansion of 52 = 5 × 5

= 25

By comparing LHS and RHS,

LHS < RHS

23 < 52

Hence, the given statement is false.

(iii) Let us consider LHS = 23 × 32

Expansion of 23 × 32= 2 × 2 × 2 × 3 × 3

= 72

Now, consider RHS = 65

Expansion of 65 = 6 × 6 × 6 × 6 × 6

= 7776

By comparing LHS and RHS,

72 ≠ 7776

LHS ≠ RHS

Hence, the given statement is false.

(iv) Let us consider LHS = 30

= 1

Now, consider RHS = 10000

= 1

By comparing LHS and RHS,

LHS = RHS

30 = 10000

Hence, the given statement is true.

 

(i) The factors of 108 = 2 × 2 × 3 × 3 × 3

= 22 × 33

The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

= 26 × 3

Then,

= (22 × 33) × (26 × 3)

= 22 + 6 × 33 + 1 … [∵am × an = am + n]

= 28 × 34

(ii) The factors of 270 = 2 × 3 × 3 × 3 × 5

= 2 × 33 × 5

(iii) The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

= 36

The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2

= 26

Then,

= (36 × 26)

= 36 × 26

(iv) The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 28 × 3

 

(i) 83 can be written as = (2 × 2 × 2)3

= (23)3

We have,

= ((25)2 × 73)/ ((23)3 × 7)

= (25 × 2 × 73)/ ((23 × 3 × 7) … [∵(am)n = amn]

= (210 × 73)/ (29 × 7)

= (210 – 9 × 73 – 1) … [∵am ÷ an = am – n]

= 2 × 72

= 2 × 7 × 7

= 98

(ii) 25 can be written as = 5 × 5

= 52

103 can be written as = 103

= (5 × 2)3

= 53 × 23

We have,

= (52 × 52 × t8)/ (53 × 23 × t4)

= (52 + 2 × t8)/ (53 × 23 × t4) … [∵am × an = am + n]

= (54 × t8)/ (53 × 23 × t4)

= (54 – 3 × t8 – 4)/ 23 … [∵am ÷ an = am – n]

= (5 × t4)/ (2 × 2 × 2)

= (5t4)/ 8

(iii) 105 can be written as = (5 × 2)5

= 55 × 25

25 can be written as = 5 × 5

= 52

65 can be written as = (2 × 3)5

= 25 × 35

Then we have,

= (35 × 55 × 25 × 52)/ (57 × 25 × 35)

= (35 × 55 + 2 × 25)/ (57 × 25 × 35) … [∵am × an = am + n]

= (35 × 57 × 25)/ (57 × 25 × 35)

= (35 – 5 × 57 – 7 × 25 – 5)

= (30 × 50 × 20) … [∵am ÷ an = am – n]

= 1 × 1 × 1

= 1

(i) Factors of 32 = 2 × 2 × 2 × 2 × 2

= 25

Factors of 4 = 2 × 2

= 22

Then,

= (23 × 34 × 22)/ (3 × 25)

= (23 + 2 × 34) / (3 × 25) … [∵am × an = am + n]

= (25 × 34) / (3 × 25)

= 25 – 5 × 34 – 1 … [∵am ÷ an = am – n]

= 20 × 33

= 1 × 33

= 33

(ii) (52)3 can be written as = (5)2 × 3 … [∵(am)n = amn]

= 56

Then,

= (56 × 54) ÷ 57

= (56 + 4) ÷ 57 … [∵am × an = am + n]

= 510 ÷ 57

= 510 – 7 … [∵am ÷ an = am – n]

= 53

(iii) (25)4 can be written as = (5 × 5)4

= (52)4

(52)4 can be written as = (5)2 × 4 … [∵(am)n = amn]

= 58

Then,

= 58 ÷ 53

= 58 – 3 … [∵am ÷ an = am – n]

= 55

(iv) Factors of 21 = 7 × 3

Then,

= (3 × 72 × 118)/ (7 × 3 × 113)

= 31-1 × 72-1 × 118 – 3

= 30 × 7 × 115

= 1 × 7 × 115

= 7 × 115

(v) = 37/ (34+3) … [∵am × an = am + n]

= 37/ 37

= 37 – 7 … [∵am ÷ an = am – n]

= 30

= 1

(vi) = 1 + 1 + 1

= 3

(vii) = 1 × 1 × 1

= 1

(viii) = (1 + 1) × 1

= (2) × 1

= 2

(ix) (4)3 can be written as = (2 × 2)3

= (22)3

(52)4 can be written as = (2)2 × 3 … [∵(am)n = amn]

= 26

Then,

= (28 × a5)/ (26 × a3)

= 28 – 6 × a5 – 3 … [∵am ÷ an = am – n]

= 22 × a2

= 2a2 … [∵(am)n = amn]

(x) = (a5 – 3) × a8 … [∵am ÷ an = am – n]

= a2 × a8

= a2 + 8 … [∵am × an = am + n]

= a10

(xi) = 45 – 5 × (a8 – 5 × b3 – 2) … [∵am ÷ an = am – n]

= 40 × (a3b)

= 1 × a3b

= a3b

(xii) = (23 + 1)2 … [∵am × an = am + n]

= (24)2

(24)2 can be written as = (2)4 × 2 … [∵(am)n = amn]

= 28

 

(a) The expanded form of the number 279404

= (2 × 100000) + (7 × 10000) + (9 × 1000) + (4 × 100) + (0 × 10) + (4 × 1)

Now, we can express it using powers of 10 in the exponent form,

(2 × 105) + (7 × 104) + (9 × 103) + (4 × 102) + (0 × 101) + (4 × 100).

(b) The expanded form of the number 3006194

= (3 × 1000000) + (0 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + 4

Now, we can express it using powers of 10 in the exponent form,

(3 × 106) + (0 × 105) + (0 × 104) + (6 × 103) + (1 × 102) + (9 × 101) + (4 × 100).

(c) The expanded form of the number 2806196

= (2 × 1000000) + (8 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + 6

Now, we can express it using powers of 10 in the exponent form,

(2 × 106) + (8 × 105) + (0 × 104) + (6 × 103) + (1 × 102) + (9 × 101) + (6 × 100).

(d) The expanded form of the number 120719

= (1 × 100000) + (2 × 10000) + (0 × 1000) + (7 × 100) + (1 × 10) + (9 × 1)

Now, we can express it using powers of 10 in the exponent form,

(1 × 105) + (2 × 104) + (0 × 103) + (7 × 102) + (1 × 101) + (9 × 100).

(e) The expanded form of the number 20068

= (2 × 10000) + (0 × 1000) + (0 × 100) + (6 × 10) + (8 × 1)

Now, we can express it using powers of 10 in the exponent form,

(2 × 104) + (0 × 103) + (0 × 102) + (6 × 101) + (8 × 100).

 

(a) The standard form of the number appearing in the given statement is 3.84 × 108m.

(b) The standard form of the number appearing in the given statement is 3 × 108m/s.

(c) The standard form of the number appearing in the given statement is 1.2756 × 107m.

(d) The standard form of the number appearing in the given statement is 1.4 × 109m.

(e) The standard form of the number appearing in the given statement is 1 × 1011.

(f) The standard form of the number appearing in the given statement is 1.2 × 1010.

(g) The standard form of the number appearing in the given statement is 3 × 1020m.

(h) The standard form of the number appearing in the given statement is 6.023 × 1022.

(i)  The standard form of the number appearing in the given statement is 1.353 × 109 cubic km.

(j) The standard form of the number appearing in the given statement is 1.027 × 109.

 

(i) The standard form of the given number is 5 × 107

(ii) The standard form of the given number is 7 × 106

(iii) The standard form of the given number is 3.1865 × 109

(iv) The standard form of the given number is 3.90878 × 105

(v) The standard form of the given number is 3.90878 × 104

(vi) The standard form of the given number is 3.90878 × 103

 

(a) The expanded form

= (8 × 10000) + (6 × 1000) + (0 × 100) + (4 × 10) + (5 × 1)

= 80000 + 6000 + 0 + 40 + 5

= 86045

(b) The expanded form

= (4 × 100000) + (0 × 10000) + (5 × 1000) + (3 × 100) + (0 × 10) + (2 × 1)

= 400000 + 0 + 5000 + 300 + 0 + 2

= 405302

(c) The expanded form

= (3 × 10000) + (0 × 1000) + (7 × 100) + (0 × 10) + (5 × 1)

= 30000 + 0 + 700 + 0 + 5

= 30705

(d) The expanded form is,

= (9 × 100000) + (0 × 10000) + (0 × 1000) + (2 × 100) + (3 × 10) + (0 × 1)

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

= 900230