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NCERT solution class 7 chapter 13 Exponents and powers 13.2 mathematics

EXERCISE 13.2


Page No 260:

Question 1:

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

(i) 32 × 34 × 38 (ii) 615 ÷ 610 (iii) a3 × a2

(iv) 7x× 72 (v)  (vi) 25 × 55

(vii) a4 × b4 (viii) (34)3

(ix)  (x) 8t ÷ 82

Answer:

(i) 32 × 34 × 38 = (3)2 + 4 + 8 (am × an = am+n)

= 314

(ii) 615 ÷ 610 = (6)15 − 10 (am ÷ an = amn)

= 65

(iii) a3 × aa(3 + 2) (am × an = am+n)

a5

(iv) 7x + 72 = 7x + 2 (am × an = am+n)

(v) (52)3 ÷ 53

= 52 × 3 ÷ 5(am)n = amn

= 56 ÷ 53

= 5(6 − 3) (am ÷ an = amn)

= 53

(vi) 25 × 55

= (2 × 5)5 [am × bm = (a × b)m]

= 105

(vii) a4 × b4

= (ab)4 [am × bm = (a × b)m]

(viii) (34)3 = 34 × 3 = 312 (am)n = amn

(ix) (220 ÷ 215) × 23

= (220 − 15) × 23 (am ÷ an = amn)

= 25 × 23

= (25 + 3) (am × an = am+n)

= 28

(x) 8t ÷ 82 = 8(t − 2) (am ÷ an = amn)

Question 2:

Simplify and express each of the following in exponential form:

(i)  (ii)  (iii) 

(iv)  (v)  (vi) 20 + 30 + 40

(vii) 20 × 30 × 40 (viii) (30 + 20) × 50 (ix) 

(x)  (xi)  (xii) 

Answer:

(i)

(ii) [(52)3 × 54] ÷ 57

= [52 × 3 × 54] ÷ 57 (am)n = amn

= [56 × 54] ÷ 57

= [56 + 4] ÷ 57 (am × an = am+n)

= 510 ÷ 57

= 510 − 7 (am ÷ an = amn)

= 53

(iii) 254 ÷ 53 = (5 ×5)4 ÷ 53

= (52)4 ÷ 53

= 52 × 4 ÷ 53 (am)n = amn

= 58 ÷ 53

= 58 − 3 (am ÷ an = amn)

= 55

(iv)

= 1 × 7 × 115 = 7 × 115

(v)

(vi) 20 + 30 + 40 = 1 + 1 + 1 = 3

(vii) 20 × 30 × 40 = 1 × 1 × 1 = 1

(viii) (30 + 20) × 50 = (1 + 1) × 1 = 2

(ix)

(x)

(xi)

(xii) (23 × 2)2 =  (am × an = am+n)

= (24)2 = 24 × 2 (am)n = amn

= 28

Question 3:

Say true or false and justify your answer:

(i) 10 × 1011 = 10011 (ii) 23 > 52

(iii) 23 × 32 = 6(iv) 30 = (1000)0

Answer:

(i) 10 × 1011 = 10011

L.H.S. = 10 × 1011 = 1011 + 1 (am × an = am+n)

= 1012

R.H.S. = 10011 = (10 ×10)11= (102)11

= 102 × 11 = 1022 (am)n = amn

As L.H.S. ≠ R.H.S.,

Therefore, the given statement is false.

(ii) 23 > 52

L.H.S. = 23 = 2 × 2 × 2 = 8

R.H.S. = 52 = 5 × 5 = 25

As 25 > 8,

Therefore, the given statement is false.

(iii) 23 × 32 = 65

L.H.S. = 23 × 32 = 2 × 2 × 2 × 3 × 3 = 72

R.H.S. = 65 = 7776

As L.H.S. ≠ R.H.S.,

Therefore, the given statement is false.

(iv) 30 = (1000)0

L.H.S. = 30 = 1

R.H.S. = (1000)0 = 1 = L.H.S.

Therefore, the given statement is true.

Page No 261:

Question 4:

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

(i) 108 × 192 (ii) 270

(iii) 729 × 64 (iv) 768

Answer:

(i) 108 × 192

= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)

= (22 × 33) × (26 × 3)

= 26 + 2 × 33 + 1 (am × an = am+n)

= 28 × 34

(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 3× 5

(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2)

= 36 × 26

(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 28 × 3

Question 5:

Simplify:

(i)  (ii)  (iii) 

Answer:

(i)

(ii)

(iii)

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