# NCERT Solutions for Class 7 Maths Exercise 13.2 Chapter 13 Exponents and Powers

NCERT Solutions for Class 7 Maths Exercise 13.2 Chapter 13 Exponents and Powers in simple PDF are available here. Laws of exponents in that multiplying and dividing powers with the same base, taking power of a power, multiplying and dividing the powers with the same exponents and miscellaneous examples using the laws of exponents are the topics covered in this exercise of NCERT Maths Solutions for Class 7 Chapter 13. While solving the exercise questions from the NCERT Class 7 book, students often face difficulty and eventually tend to pile up their doubts. These Solutions are prepared by our expert tutors with step by step explanations.

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

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

(i) 32 Ã— 34 Ã— 38

Solution:-

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

Then,

= (3)2 + 4 + 8

= 314

(ii) 615 Ã· 610

Solution:-

By the rule of dividing the powers with same base = am Ã· an = am â€“ n

Then,

= (6)15 â€“ 10

= 65

(iii) a3 Ã— a2

Solution:-

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

Then,

= (a)3 + 2

= a5

(iv) 7x Ã— 72

Solution:-

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

Then,

= (7)x + 2

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(v) (52)3 Ã· 53

Solution:-

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

(52)3 can be written as = (5)2 Ã— 3

= 56

Now, 56 Ã· 53

By the rule of dividing the powers with same base = am Ã· an = am â€“ n

Then,

= (5)6 â€“ 3

= 53

(vi) 25 Ã— 55

Solution:-

By the rule of multiplying the powers with same exponents = am Ã— bm = abm

Then,

= (2 Ã— 5)5

= 105

(vii) a4 Ã— b4

Solution:-

By the rule of multiplying the powers with same exponents = am Ã— bm = abm

Then,

= (a Ã— b)4

= ab4

(viii) (34)3

Solution:-

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

(34)3 can be written as = (3)4 Ã— 3

= 312

(ix) (220 Ã· 215) Ã— 23

Solution:-

By the rule of dividing the powers with same base = am Ã· an = am â€“ n

(220 Ã· 215) can be simplified as,

= (2)20 â€“ 15

= 25

Then,

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

25 Ã— 23 can be simplified as,

= (2)5 + 3

= 28

(x) 8t Ã· 82

Solution:-

By the rule of dividing the powers with same base = am Ã· an = am â€“ n

Then,

= (8)t â€“ 2

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

(i) (23 Ã— 34 Ã— 4)/ (3 Ã— 32)

Solution:-

Factors of 32 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

= 25

Factors of 4 = 2 Ã— 2

= 22

Then,

= (23 Ã— 34 Ã— 22)/ (3 Ã— 25)

= (23 + 2 Ã— 34) / (3 Ã— 25) â€¦ [âˆµam Ã— an = am + n]

= (25 Ã— 34) / (3 Ã— 25)

= 25 â€“ 5 Ã— 34 â€“ 1 â€¦ [âˆµam Ã· an = am â€“ n]

= 20 Ã— 33

= 1 Ã— 33

= 33

(ii) ((52)3 Ã— 54) Ã· 57

Solution:-

(52)3 can be written as = (5)2 Ã— 3 â€¦ [âˆµ(am)n = amn]

= 56

Then,

= (56 Ã— 54) Ã· 57

= (56 + 4) Ã· 57 â€¦ [âˆµam Ã— an = am + n]

= 510 Ã· 57

= 510 â€“ 7 â€¦ [âˆµam Ã· an = am â€“ n]

= 53

(iii) 254 Ã· 53

Solution:-

(25)4 can be written as = (5 Ã— 5)4

= (52)4

(52)4 can be written as = (5)2 Ã— 4 â€¦ [âˆµ(am)n = amn]

= 58

Then,

= 58 Ã· 53

= 58 â€“ 3 â€¦ [âˆµam Ã· an = am â€“ n]

= 55

(iv) (3 Ã— 72 Ã— 118)/ (21 Ã— 113)

Solution:-

Factors of 21 = 7 Ã— 3

Then,

= (3 Ã— 72 Ã— 118)/ (7 Ã— 3 Ã— 113)

= 31-1 Ã— 72-1 Ã— 118 â€“ 3

= 30 Ã— 7 Ã— 115

= 1 Ã— 7 Ã— 115

= 7 Ã— 115

(v) 37/ (34 Ã— 33)

Solution:-

= 37/ (34+3) â€¦ [âˆµam Ã— an = am + n]

= 37/ 37

= 37 â€“ 7 â€¦ [âˆµam Ã· an = am â€“ n]

= 30

= 1

(vi) 20 + 30 + 40

Solution:-

= 1 + 1 + 1

= 3

(vii) 20 Ã— 30 Ã— 40

Solution:-

= 1 Ã— 1 Ã— 1

= 1

(viii) (30 + 20) Ã— 50

Solution:-

= (1 + 1) Ã— 1

= (2) Ã— 1

= 2

(ix) (28 Ã— a5)/ (43 Ã— a3)

Solution:-

(4)3 can be written as = (2 Ã— 2)3

= (22)3

(52)4 can be written as = (2)2 Ã— 3 â€¦ [âˆµ(am)n = amn]

= 26

Then,

= (28 Ã— a5)/ (26 Ã— a3)

= 28 â€“ 6 Ã— a5 â€“ 3 â€¦ [âˆµam Ã· an = am â€“ n]

= 22 Ã— a2

= 2a2 â€¦ [âˆµ(am)n = amn]

(x) (a5/a3) Ã— a8

Solution:-

= (a5 â€“ 3) Ã— a8 â€¦ [âˆµam Ã· an = am â€“ n]

= a2 Ã— a8

= a2 + 8 â€¦ [âˆµam Ã— an = am + n]

= a10

(xi) (45 Ã— a8b3)/ (45 Ã— a5b2)

Solution:-

= 45 â€“ 5 Ã— (a8 â€“ 5 Ã— b3 â€“ 2) â€¦ [âˆµam Ã· an = am â€“ n]

= 40 Ã— (a3b)

= 1 Ã— a3b

= a3b

(xii) (23 Ã— 2)2

Solution:-

= (23 + 1)2 â€¦ [âˆµam Ã— an = am + n]

= (24)2

(24)2 can be written as = (2)4 Ã— 2 â€¦ [âˆµ(am)n = amn]

= 28

3. Say true or false and justify your answer:

(i) 10 Ã— 1011 = 10011

Solution:-

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

= 101 + 11 â€¦ [âˆµam Ã— an = am + n]

= 1012

Now, consider Right Hand Side (RHS) = 10011

= (10 Ã— 10)11

= (101 + 1)11

= (102)11

= (10)2 Ã— 11 â€¦ [âˆµ(am)n = amn]

= 1022

By comparing LHS and RHS,

LHS â‰  RHS

Hence, the given statement is false.

(ii) 23 > 52

Solution:-

Let us consider LHS = 23

Expansion of 23 = 2 Ã— 2 Ã— 2

= 8

Now, consider RHS = 52

Expansion of 52 = 5 Ã— 5

= 25

By comparing LHS and RHS,

LHS < RHS

23 < 52

Hence, the given statement is false.

(iii) 23 Ã— 32 = 65

Solution:-

Let us consider LHS = 23 Ã— 32

Expansion of 23 Ã— 32= 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3

= 72

Now, consider RHS = 65

Expansion of 65 = 6 Ã— 6 Ã— 6 Ã— 6 Ã— 6

= 7776

By comparing LHS and RHS,

72Â â‰  7776

LHSÂ â‰  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 Ã— 192

Solution:-

The factors of 108 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3

= 22 Ã— 33

The factors of 192 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3

= 26 Ã— 3

Then,

= (22 Ã— 33) Ã— (26 Ã— 3)

= 22 + 6 Ã— 33 + 1 â€¦ [âˆµam Ã— an = am + n]

= 28 Ã— 34

(ii) 270

Solution:-

The factors of 270 = 2 Ã— 3 Ã— 3 Ã— 3 Ã— 5

= 2 Ã— 33 Ã— 5

(iii) 729 Ã— 64

The factors of 729 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

= 36

The factors of 64 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

= 26

Then,

= (36 Ã— 26)

= 36 Ã— 26

(iv) 768

Solution:-

The factors of 768 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3

= 28 Ã— 3

5. Simplify:

(i) ((25)2 Ã— 73)/ (83 Ã— 7)

Solution:-

83 can be written as = (2 Ã— 2 Ã— 2)3

= (23)3

We have,

= ((25)2 Ã— 73)/ ((23)3 Ã— 7)

= (25 Ã— 2 Ã— 73)/ ((23 Ã— 3 Ã— 7) â€¦ [âˆµ(am)n = amn]

= (210 Ã— 73)/ (29 Ã— 7)

= (210 â€“ 9 Ã— 73 â€“ 1) â€¦ [âˆµam Ã· an = am â€“ n]

= 2 Ã— 72

= 2 Ã— 7 Ã— 7

= 98

(ii) (25 Ã— 52 Ã— t8)/ (103 Ã— t4)

Solution:-

25 can be written as = 5 Ã— 5

= 52

103 can be written as = 103

= (5 Ã— 2)3

= 53 Ã— 23

We have,

= (52 Ã— 52 Ã— t8)/ (53 Ã— 23 Ã— t4)

= (52 + 2 Ã— t8)/ (53 Ã— 23 Ã— t4) â€¦ [âˆµam Ã— an = am + n]

= (54 Ã— t8)/ (53 Ã— 23 Ã— t4)

= (54 â€“ 3 Ã— t8 â€“ 4)/ 23 â€¦ [âˆµam Ã· an = am â€“ n]

= (5 Ã— t4)/ (2 Ã— 2 Ã— 2)

= (5t4)/ 8

(iii) (35 Ã— 105 Ã— 25)/ (57 Ã— 65)

Solution:-

105 can be written as = (5 Ã— 2)5

= 55 Ã— 25

25 can be written as = 5 Ã— 5

= 52

65 can be written as = (2 Ã— 3)5

= 25 Ã— 35

Then we have,

= (35 Ã— 55 Ã— 25 Ã— 52)/ (57 Ã— 25 Ã— 35)

= (35 Ã— 55 + 2 Ã— 25)/ (57 Ã— 25 Ã— 35) â€¦ [âˆµam Ã— an = am + n]

= (35 Ã— 57 Ã— 25)/ (57 Ã— 25 Ã— 35)

= (35 â€“ 5 Ã— 57 â€“ 7 Ã— 25 â€“ 5)

= (30 Ã— 50 Ã— 20) â€¦ [âˆµam Ã· an = am â€“ n]

= 1 Ã— 1 Ã— 1

= 1

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