Gujarat Board 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) 3² x 3⁴ x 3⁸
(ii) 6¹⁵ ÷ 6¹⁰
(iii) a³ x a²
(iv) 7^x x 7²
(v) (5²)³ ÷ 5³
(vi) 2⁵ x 5⁵
(vii) a⁴ x b⁴
(viii) (3⁴)³
(ix) (2²⁰ ÷ 2¹⁵) x 2³
(x) 8^t ÷ 8²
Solution:
(i) 3² x 3⁴ x 3⁸
We have: 3² x 3⁴ x 3⁸ = 3²⁺⁴⁺⁸ (∵ a^m x aⁿ x a^r = a^m+n+r)
= 314
Thus, 3² x 3⁴ x 3⁸ = 314

(ii) 6¹⁵ ÷ 6¹⁰
We have: 6¹⁵ ÷ 6¹⁰ = 6^15-10 = 6⁵ (∵ a^m ÷ aⁿ = a^m-n )
Thus, 6¹⁵ ÷ 6¹⁰ = 6⁵

(iii) a³ x a²
We have: a³ x a² = a³⁺² = a⁵ (∵ a^m x aⁿ = a^m+n)
Thus, a³ x a²= a⁵

(iv) 7^x x 7²:
We have: 7^x x 7² = 7^x+2 (∵ a^m x aⁿ = a^m+n)
Thus, 7^x x 7²= 7^x+2

(v) (5²)³ .÷ 5³:
We have: (5²)³ + 5³ = 5^2×3 ÷ + 5³
∵ [ (a^m)ⁿ = a^mn ]
= 5⁶ ÷ 5³
5^6-3 = 5³ (∵ a^m ÷ aⁿ = a^m-n)
Thus, (5²)³ ÷ 5³ = 53

(vi) 2⁵x 5⁵:
We have: 2⁵ x 5⁵ = (2 x 5)⁵ = (10)⁵ (∵ a^m x aⁿ = (ab)^m)
Thus, 2⁵ x 5⁵ = 10⁵

(vii) a⁴ x b⁴:
We have: a⁴ x b⁴ = (ab)⁵ [ ∵ a^m x aⁿ = (ab)^m ]
Thus, a⁴ x b⁴ = (ab)⁴

(viii) (3⁴)³:
We have: (3⁴)³ = 3^4×3 = 3¹² [ ∵ a^m x aⁿ = (ab)^m ]
Thus, (3⁴)³ = 312

(ix) (2²⁰ ÷ 2¹⁵) x 2³:
We have: (2²⁰ ÷ 2¹⁵) x 2³ (∵ a^m ÷ aⁿ = a^m-n)
= (2⁵) x 2³
= 2⁵ x 2³
= 2⁵⁺³ (∵ a^m x aⁿ = a^m+n)
= 2⁸
= 2⁸
Thus, (2²⁰ + 2¹⁵) x 2³ = 2⁸

(x) 8^t ÷ 8²:
We have:
8^t ÷ 8² = 8^t-2 (∵ a^m ÷ aⁿ = a^m-n)
Thus, 8^t ÷ 8² = 8^t-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) [(5²)³ x 5⁴] ÷ 5⁷
(iii) 25⁴ ÷ 5³
(iv) (frac{3 times 7^{2} times 11^{8}}{21 times 11^{3}})
(v) (frac{3^{7}}{3^{4} times 3^{3}})
(vi) 2° + 3° + 4°
(vii) 2° x 3° x 4°
(viii) (3° + 2°) x 5°
(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) (2³ x 2)²
Solution:

(vi) 2° + 3° + 4°:
Since a° = 1
∴ 2° = 1, 3° = 1 and 4° = 1
We have : 2° + 3° + 4° = 1 + 1 + 1 = 3

(vii) 2° x 3° x 4°:
Since a° = 1
∴ 2° = 1, 3° = 1 and 4° = 1
Now, 2° x 3° x 4° = 1 x 1 x 1 = 1

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

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

(xii) (2³ x 2)²:
We have: (2³ x 2)² = (2³⁺¹ )² (∵ a^m x aⁿ = a^m )
= (2⁴)²
= 2⁴ x 2 (∵ (a^m)ⁿ = a^mn )
= 2⁸

Question 3.
Say true or false and justify your answer:
(i) 10 x 10¹¹ = 100¹¹
(ii) 2³ > 5²
(iii) 2³ x 3² = 65
(iv) 3° = (1000)°
Solution:
(i) 10 x 10¹¹ = 100¹¹
∵ 10 x 10¹¹ = 10¹ + 11 = 10¹²
But 10¹² ≠ 100¹¹
∴ 10 x 10¹¹ ≠ 100¹¹
i.e. The statement 10 x 10¹¹ = 100¹¹ is false.

(ii) 2³ > 5²
∵ 2³ = 2 x 2 x 2 = 8
5² = 5 x 5 = 25
and 8 < 25, i.e. 2² < 5²
Thus, the statement 2³ > 5² is false.

(iii) 2³ x 3² = 6⁵
∵ 2³ x 3² = 2 x 2 x 2 x 3 x 3 = 72
and 65 = 6 x 6 x 6 x 6 x 6 = 7776
Also 72 ≠ 7776
∴ 2³ x 3² ≠ 6⁵
i.e. The statement 2³ x 3² = 65 is false

(iv) 3° = (1000)°
Since, 3° = 1 and (1000)° = 1
∴ 3° = (1000)° 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:

∵ 108 = 2 x 2 x 31 x 3 x 3
192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
∴ 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 3³ x 5¹
= 2 x 3³ x 5
Thus, 270 = 2 x 3³ 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

∴ 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: