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# Mathematics Class 7 – Chapter 13 – Exponents and Powers – Exercise 13.2 – NCERT Exercise Solution

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

(i) 32 × 34 × 38

Solution:

We use the rule of multiply for the power with same base am x an= am+n

= (3) 2+3+8

= 314

(ii) 615 ÷ 610

Solution:

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

= (6) 15-10

= 65

(iii) a3 × a2

Solution:

= (a) 3+2

= a5

(iv) 7x × 72

Solution:

= (7) x+2

(v) (52) 3 ÷ 53

Solution:

(52) 3 can be written as = (5) 2×3

= 56

Now,

56 ÷ 53

= (5) 6-3

= 53

(vi) 25 × 55

Solution:

= (2 × 5) 5

= 105

(vii) a4 × b4

Solution:

= (a × b) 4

= (ab)4

(viii) (34) 3

Solution:

=(34) 3 can be written as = (3) 4×3

= 312

(ix) (220 ÷ 215) × 23

Solution:

(220 ÷ 215) can be simplified as,

= (220 – 15)

= 25

Then,

25 × 23 can be simplified as,

= (2)5 + 3

= 28

(x) 8t ÷ 82

Solution:

= (8)t – 2

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

(i) (23 × 34 × 4)/ (3 × 32)

Solution:

Factors of 4 = 2 × 2

= 22

Factors of 32 = 2 × 2 × 2 × 2 × 2

= 25

Then,

= (23 × 34 × 22)/ (3 × 25)

[we know that, am × an = am+n]

= (25 × 34) / (3 × 25)

= 25-5 × 34-1 … [by: am ÷ an = am-n]

= 20 × 33

= 1 × 33

= 33

(ii) ((52)3 × 54) ÷ 57

Solution:

(52)3 can be written as = (5)2×3

= 56

Now,

= (56 × 54) ÷ 57

= (56+4) ÷ 57

= 510 ÷ 57

= 510-7

= 53

(iii) 254 ÷ 53

Solution:

(25)4 can be written as = (52)4

(52)4 can be written as = (5)2×4

= 58

Then,

= 58 ÷ 53

= 58-3

= 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 × 71 × 115

= 1 × 7 × 115

= 7 × 115

(v) 37/ (34 × 33)

Solution:

= 37/ (34+3)

= 37/ 37

= 37-7

= 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 = (22) 3

(52) 4 can be written as = (2) 2×3

= 26

Now,

= (28 × a5)/ (26 × a3)

= 28-6 × a5-3

= 22 × a2

= 2a2

(x) (a5/a3) × a8

Solution:

= (a5-3) × a8

= a2 × a8

= a10

(xi) (45 × a8b3)/ (45 × a5b2)

Solution:

= 45-5 × (a8-5 × b3-2)

= 40 × (a3b1)

= 1 × a3b

= a3b

(xii) (23 × 2) 2

Solution:

= (23+1) 2

= (24) 2

= 28

(i) 10 × 1011 = 10011

Solution:

According to question:

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

= 101+11

= 1012

Now,

Right Hand Side (RHS) = 10011

= (10 × 10) 11

= (101+1) 11

= (102) 11

= 1022

Now,

By comparing LHS and RHS,

LHS ≠ RHS

Hence, the given statement is false.

(ii) 23 > 52

Solution:

According t question:

Let us consider LHS = 23

23 = 8

Now,

consider RHS = 52

52 = 25

By comparing LHS and RHS,

LHS < RHS

23 < 52

Hence, the given statement is false.

(iii) 23 × 32 = 65

Solution:

According to question:

Let us consider LHS = 23 × 32

23 × 32 = 2 × 2 × 2 × 3 × 3

= 72

Now,

consider RHS = 65

65 = 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:

According to question:

The factors of 108 = 2 × 2 × 3 × 3 × 3

= 22 × 33

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

= 26 × 3

Now,

= (22 × 33) × (26 × 3)

= 22+6 × 33+1

= 28 × 34

(ii) 270

Solution:

According to question:

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

= 2 × 33 × 5

(iii) 729 × 64

Solution:

According to question:

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36

The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2 = 26

Now,

= 36 × 26

(iv) 768

Solution:

According to question:

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

= 28 × 3

5. Simplify:

(i) ((25)2 × 73)/ (83 × 7)

Solution:

We can write 83 as = (2 × 2 × 2)3

= (23)3

Now we have,

= ((25)2 × 73)/ ((23) 3 × 7)

= (210× 73)/ ((29 × 7)

Now,

= (210-9 × 73-1)

= 2 × 72

= 98

(ii) (25 × 52 × t8)/ (103 × t4)

Solution:

We can write 25 as = 5 × 5 = 52

103 can be written as = (5 × 2) 3

= 53 × 23

Now we have,

= (52 × 52 × t8)/ (53 × 23 × t4)

= (52+2 × t8)/ (53 × 23 × t4)

Now,

= (54 × t8)/ (53 × 23 × t4)

= (54-3 × t8-4)/ 23

= (5t4)/ 8

(iii) (35 × 105 × 25)/ (57 × 65)

Solution:

We can write 105 as = (5 × 2) 5

= 55 × 25

We can write 25 as = 5 × 5 = 52

We can write 65 as = (2 × 3) 5

= 25 × 35

Now we have,

= (35 × 55 × 25 × 52)/ (57 × 25 × 35)

= (35 × 55+2 × 25)/ (57 × 25 × 35)

= (35 × 57 × 25)/ (55 × 25 × 35)

Now,

= (35-5 × 57-7 × 25-5)

= (30 × 50 × 20)

= 1 × 1 × 1

= 1

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