# GSEB Solutions Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2

Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2

Question 1.
Using laws of exponents, simplify and write the answer in exponential form:
(i) 3Ā² x 34 x 38
(ii) 615 Ć· 610
(iii) a3 x a2
(iv) 7x x 72
(v) (52)3 Ć· 53
(vi) 25 x 55
(vii) a4 x b4
(viii) (34)3
(ix) (220 Ć· 215) x 23
(x) 8t Ć· 82
Solution:
(i) 3Ā² x 34 x 38
We have: 3Ā² x 34 x 38 = 32+4+8 (āµ am x an x ar = am+n+r)
= 314
Thus, 32 x 34 x 38 = 314

(ii) 615 Ć· 610
We have: 615 Ć· 610 = 615-10 = 65 (āµ am Ć· an = am-n )
Thus, 615 Ć· 610 = 65

(iii) a3 x a2
We have: a3 x a2 = a3+2 = a5 (āµ am x an = am+n)
Thus, a3 x a2= a5

(iv) 7x x 72:
We have: 7x x 72 = 7x+2 (āµ am x an = am+n)
Thus, 7x x 72= 7x+2

(v) (5Ā²)3 .Ć· 53:
We have: (52)3 + 53 = 52Ć3 Ć· + 53
āµ [ (am)n = amn ]
= 56 Ć· 53
56-3 = 53 (āµ am Ć· an = am-n)
Thus, (5Ā²)Ā³ Ć· 5Ā³ = 53

(vi) 25x 55:
We have: 25 x 55 = (2 x 5)5 = (10)5 (āµ am x an = (ab)m)
Thus, 25 x 55 = 105

(vii) a4 x b4:
We have: a4 x b4 = (ab)5 [ āµ am x an = (ab)m ]
Thus, a4 x b4 = (ab)4

(viii) (34)3:
We have: (34)3 = 34Ć3 = 312 [ āµ am x an = (ab)m ]
Thus, (34)3 = 312

(ix) (220 Ć· 215) x 23:
We have: (220 Ć· 215) x 23 (āµ am Ć· an = am-n)
= (25) x 23
= 25 x 23
= 25+3 (āµ am x an = am+n)
= 28
= 28
Thus, (220 + 215) x 23 = 28

(x) 8t Ć· 82:
We have:
8t Ć· 82 = 8t-2 (āµ am Ć· an = am-n)
Thus, 8t Ć· 82 = 8t-2

Question 2.
Simplify and express each of the following in exponential form:
(i) $$\frac{2^{3} \times 3^{4} \times 4}{3 \times 32}$$
(ii) [(5Ā²)Ā³ x 54] Ć· 57
(iii) 254 Ć· 53
(iv) $$\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}$$
(v) $$\frac{3^{7}}{3^{4} \times 3^{3}}$$
(vi) 2Ā° + 3Ā° + 4Ā°
(vii) 2Ā° x 3Ā° x 4Ā°
(viii) (3Ā° + 2Ā°) x 5Ā°
(ix) $$\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}$$
(x) $$\left[\frac{a^{5}}{a^{3}}\right]$$
(xi) $$\frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}}$$
(xii) (2Ā³ x 2)Ā²
Solution:

(vi) 2Ā° + 3Ā° + 4Ā°:
Since aĀ° = 1
ā“ 2Ā° = 1, 3Ā° = 1 and 4Ā° = 1
We have : 2Ā° + 3Ā° + 4Ā° = 1 + 1 + 1 = 3

(vii) 2Ā° x 3Ā° x 4Ā°:
Since aĀ° = 1
ā“ 2Ā° = 1, 3Ā° = 1 and 4Ā° = 1
Now, 2Ā° x 3Ā° x 4Ā° = 1 x 1 x 1 = 1

(viii) (3Ā° + 2Ā°) x 5Ā°:
We have oĀ° = 1
(3Ā° + 2Ā°) x 5Ā° = (1 + 1) x 1
= 2 x 1 = 2

(ix) $$\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}$$

(xii) (2Ā³ x 2)Ā²:
We have: (2Ā³ x 2)Ā² = (23+1 )Ā² (āµ am x an = am )
= (24)Ā²
= 24 x 2 (āµ (am)n = amn )
= 28

Question 3.
(i) 10 x 1011 = 10011
(ii) 23 > 5Ā²
(iii) 2Ā³ x 3Ā² = 65
(iv) 3Ā° = (1000)Ā°
Solution:
(i) 10 x 1011 = 10011
āµ 10 x 1011 = 101 + 11 = 1012
But 1012 ā  10011
ā“ 10 x 1011 ā  10011
i.e. The statement 10 x 1011 = 10011 is false.

(ii) 23 > 52
āµ 23 = 2 x 2 x 2 = 8
5Ā² = 5 x 5 = 25
and 8 < 25, i.e. 2Ā² < 5Ā²
Thus, the statement 23 > 5Ā² is false.

(iii) 2Ā³ x 3Ā² = 65
āµ 2Ā³ x 3Ā² = 2 x 2 x 2 x 3 x 3 = 72
and 65 = 6 x 6 x 6 x 6 x 6 = 7776
Also 72 ā  7776
ā“ 2Ā³ x 3Ā² ā  65
i.e. The statement 2Ā³ x 3Ā² = 65 is false

(iv) 3Ā° = (1000)Ā°
Since, 3Ā° = 1 and (1000)Ā° = 1
ā“ 3Ā° = (1000)Ā° is a true statement.

Question 4.
Express each of the following as a product of prime factors only in exponential form:
(i) 108 x 192
(ii) 270
(iii) 729 x 64
(iv) 768
Solution:
(i) 108 x 192:

āµ 108 = 2 x 2 x 31 x 3 x 3
192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
ā“ 108 x 192 = (2 x 2 x 3 x 3 x 3) x (2 x 2 x 2 x 2 x 2 x 2 x 3)
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3
= 28 x 34
Thus, 108 x 192 = 28 x 34

(ii) 270:

We have: 270 = 2 x 3 x 3 x 3 x 5
= 21 x 3Ā³ x 51
= 2 x 3Ā³ x 5
Thus, 270 = 2 x 3Ā³ x 5

(iii) 729 x 64:
We have: 729 = 3 x 3 x 3 x 3 x 3 x 3
64 = 2 x 2 x 2 x 2 x 2 x 2

ā“ 729 x 64 = (3 x 3 x 3 x 3 x 3 x 3) x (2 x 2 x 2 x 2 x 2 x 2)
= 36 x 26
Thus, 729 x 64 = 36 x 26

(iv) 768:

We have: 768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
= 28 x 31
= 28 x 3
Thus, 768 = 28 x 3

Question 5.
Simplify:
(i) $$\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}$$
(ii) $$\frac{25 \times 5^{2} \times t^{8}}{10^{3} \times t^{4}}$$
(iii) $$\frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}}$$
Solution: