## Access answers to Maths NCERT Solutions for Class 7 Chapter 13 – Exponents and Powers Exercise 13.2

### Access answers to Maths NCERT Solutions for Class 7 Chapter 13 – Exponents and Powers Exercise 13.2

1. Using laws of exponents, simplify and write the answer in exponential form:

(i) 32 × 34 × 38

Solution:-

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

Then,

= (3)2 + 4 + 8

= 314

(ii) 615 ÷ 610

Solution:-

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

Then,

= (6)15 – 10

= 65

(iii) a3 × a2

Solution:-

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

Then,

= (a)3 + 2

= a5

(iv) 7x × 72

Solution:-

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

Then,

= (7)x + 2

(v) (52)3 ÷ 53

Solution:-

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

(52)3 can be written as = (5)2 × 3

= 56

Now, 5÷ 53

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

Then,

= (5)6 – 3

= 53

(vi) 25 × 55

Solution:-

By the rule of multiplying the powers with same exponents = a× bm = abm

Then,

= (2 × 5)5

= 105

(vii) a4 × b4

Solution:-

By the rule of multiplying the powers with same exponents = a× bm = abm

Then,

= (a × b)4

= ab4

(viii) (34)3

Solution:-

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

(34)3 can be written as = (3)4 × 3

= 312

(ix) (220 ÷ 215) × 23

Solution:-

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

(220 ÷ 215) can be simplified as,

= (2)20 – 15

= 25

Then,

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

25 × 23 can be simplified as,

= (2)5 + 3

= 28

(x) 8t ÷ 82

Solution:-

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

Then,

= (8)t – 2

2. Simplify and express each of the following in exponential form:

(i) (23 × 34 × 4)/ (3 × 32)

Solution:-

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) … [∵a× an = am + n]

= (25 × 34) / (3 × 25)

= 25 – 5 × 34 – 1 … [∵a÷ an = am – n]

= 20 × 33

= 1 × 33

= 33

(ii) ((52)3 × 54) ÷ 57

Solution:-

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

= 56

Then,

= (5× 54) ÷ 57

= (56 + 4) ÷ 57 … [∵a× an = am + n]

= 510 ÷ 57

= 510 – 7 … [∵a÷ an = am – n]

= 53

(iii) 254 ÷ 53

Solution:-

(25)4 can be written as = (5 × 5)4

= (52)4

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

= 58

Then,

= 58 ÷ 53

= 58 – 3 … [∵a÷ an = am – n]

= 55

(iv) (3 × 72 × 118)/ (21 × 113)

Solution:-

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 × 33)

Solution:-

= 37/ (34+3) … [∵a× an = am + n]

= 37/ 37

= 37 – 7 … [∵a÷ an = am – n]

= 30

= 1

(vi) 20 + 30 + 40

Solution:-

= 1 + 1 + 1

= 3

(vii) 2× 30 × 40

Solution:-

= 1 × 1 × 1

= 1

(viii) (30 + 20) × 50

Solution:-

= (1 + 1) × 1

= (2) × 1

= 2

(ix) (28 × a5)/ (43 × a3)

Solution:-

(4)3 can be written as = (2 × 2)3

= (22)3

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

= 26

Then,

= (28 × a5)/ (26 × a3)

= 28 – 6 × a5 – 3 … [∵a÷ an = am – n]

= 2× a2

= 2a2 … [∵(am)= amn]

(x) (a5/a3) × a8

Solution:-

= (a5 – 3) × a8 … [∵a÷ an = am – n]

= a2 × a8

= a2 + 8 … [∵a× an = am + n]

= a10

(xi) (45 × a8b3)/ (45 × a5b2)

Solution:-

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

= 40 × (a3b)

= 1 × a3b

= a3b

(xii) (23 × 2)2

Solution:-

= (23 + 1)2 … [∵a× an = am + n]

= (24)2

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

= 28

3. Say true or false and justify your answer:

(i) 10 × 1011 = 10011

Solution:-

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

= 101 + 11 … [∵a× an = am + n]

= 1012

Now, consider Right Hand Side (RHS) = 10011

= (10 × 10)11

= (101 + 1)11

= (102)11

= (10)2 × 11 … [∵(am)= amn]

= 1022

By comparing LHS and RHS,

LHS ≠ RHS

Hence, the given statement is false.

(ii) 23 > 52

Solution:-

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) 23 × 32 = 65

Solution:-

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) 30 = (1000)0

Solution:-

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.

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

(i) 108 × 192

Solution:-

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 … [∵a× an = am + n]

= 2× 34

(ii) 270

Solution:-

The factors of 270 = 2 × 3 × 3 × 3 × 5

= 2 × 33 × 5

(iii) 729 × 64

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) 768

Solution:-

The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 28 × 3

5. Simplify:

(i) ((25)2 × 73)/ (83 × 7)

Solution:-

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)= amn]

= (210 × 73)/ (29 × 7)

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

= 2 × 72

= 2 × 7 × 7

= 98

(ii) (25 × 52 × t8)/ (103 × t4)

Solution:-

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) … [∵a× an = am + n]

= (54 × t8)/ (53 × 23 × t4)

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

= (5 × t4)/ (2 × 2 × 2)

= (5t4)/ 8

(iii) (35 × 105 × 25)/ (57 × 65)

Solution:-

10can 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) … [∵a× an = am + n]

= (35 × 57 × 25)/ (57 × 25 × 35)

= (35 – 5 × 57 – 7 × 25 – 5)

= (30 × 50 × 20) … [∵a÷ an = am – n]

= 1 × 1 × 1

= 1