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.

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 13 Exponents and Powers – Exercise 13.2

 

ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2

 

Access other exercises of NCERT Solutions For Class 7 Chapter 13 – Exponents and Powers

Exercise 13.1 Solutions

Exercise 13.3 Solutions

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 = am × an = am + n

Then,

= (3)2 + 4 + 8

= 314

(ii) 615 ÷ 610

Solution:-

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

Then,

= (6)15 – 10

= 65

(iii) a3 × a2

Solution:-

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

Then,

= (a)3 + 2

= a5

(iv) 7x × 72

Solution:-

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

Then,

= (7)x + 2

 

(v) (52)3 ÷ 53

Solution:-

By the rule of taking 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) 25 × 55

Solution:-

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

Then,

= (2 × 5)5

= 105

(vii) a4 × b4

Solution:-

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

Then,

= (a × 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 × 3

= 312

(ix) (220 ÷ 215) × 23

Solution:-

By the rule of dividing the powers with 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 same base = am × 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 = am ÷ 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) … [∵am × an = am + n]

= (25 × 34) / (3 × 25)

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

= 20 × 33

= 1 × 33

= 33

(ii) ((52)3 × 54) ÷ 57

Solution:-

(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) 254 ÷ 53

Solution:-

(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) (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) … [∵am × an = am + n]

= 37/ 37

= 37 – 7 … [∵am ÷ an = am – n]

= 30

= 1

(vi) 20 + 30 + 40

Solution:-

= 1 + 1 + 1

= 3

(vii) 20 × 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)n = amn]

= 26

Then,

= (28 × a5)/ (26 × a3)

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

= 22 × a2

= 2a2 … [∵(am)n = amn]

(x) (a5/a3) × a8

Solution:-

= (a5 – 3) × a8 … [∵am ÷ an = am – n]

= a2 × a8

= a2 + 8 … [∵am × an = am + n]

= a10

(xi) (45 × a8b3)/ (45 × a5b2)

Solution:-

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

= 40 × (a3b)

= 1 × a3b

= a3b

(xii) (23 × 2)2

Solution:-

= (23 + 1)2 … [∵am × an = am + n]

= (24)2

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

= 28 × 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)n = amn]

= (210 × 73)/ (29 × 7)

= (210 – 9 × 73 – 1) … [∵am ÷ 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) … [∵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) (35 × 105 × 25)/ (57 × 65)

Solution:-

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


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