KSEEB Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2

Students can Download Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus 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 × 3⁴ × 3⁸
(ii) 6¹⁵ ÷ 6¹⁰
(iii) a³ × a²
(iv) 7^x × 7²
(v) (5²)³ ÷ 5²
(vi) 2⁵ × 5⁵
(vii) (a⁴ × b⁴)
(viii) (3⁴)³
(ix) (2²⁰ ÷ 2¹⁵) × 2³
(x) 8^t ÷ 8²
Solution:
(i) 3² × 3⁴ × 3⁸ = 3²⁺⁴⁺⁸ = 3¹⁴
[using the law a^m × aⁿ = a^m+n

(ii) 6¹⁵ ÷ 6¹⁰ = 6^15-10 = 6⁵
[using the law a^m ÷ aⁿ =a^m+n]

(iii) a³ × a² = a³⁺² = a⁵
[using the law a^m x aⁿ = a^m+n]

(iv) 7^x × 7² = 7^x+2 = 7^x+2
[using the law a^m x aⁿ = a^m+n]

(v) (52)³ ÷ 5³ = 5^2×3 + 5³
[using the law (a^m)ⁿ = a^mn] = 5⁶ ÷ 5³
[using the law a^m ÷ aⁿ = a^m+n] = 5^6-3 = 5³

(vi) 2⁵ × 5⁵ = (2 × 5)⁵ – 10⁵
[using the law a^m × a^m = (ab)^m]

(vii) a⁴ × b⁴ = (a × b)⁴ = (ab)⁴
[using the law (a^m × b^m) (ab)^m

(viii) (3⁴)³ = 3^4×3 = 3¹² [using (a^m)ⁿ = a^mn ]

(ix) (2²⁰ ÷ 2¹⁵) × 2³
2^20-15 × 2³ = 2⁵ × 2³
[using the law a^m ÷ aⁿ = a^m-n
2⁵ × 2³ = 2⁵⁺³ = 28
[using the law a^m × a^m = a^m+n
(x) 8^t ÷ 8²
8^t-2 = 8^t-2 [using a^m ÷ aⁿ = a^m-n

Question 2.
Simplify and express each of the following in exponential form:
(i) \(\frac{2^{3} \times 3^{4} \times 4}{3 \times 32}\)
Solution:

(ii) ((5²)³) × 5⁴) ÷ 5⁷
(5⁶ × 5⁴) ÷ 5⁷ ⇒ 5⁶⁺⁴ ÷ 5⁷
= 5¹⁰ ÷ 5⁷ = 5^10-7 = 5³
Hence [(5²)³ × 5⁴] + 5⁷ = 5³
25 = 5²

(iii) 25⁴ ÷ 5³
(5²)⁴ ÷ 5³
= 5⁸ + 5³= 5^8-3 = 5⁵
Hence 25⁴ ÷ 5³ = 5⁵

(iv)\(\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}\)
Solution:

(v) \(\frac{3^{7}}{3^{4} \times 3^{3}}\)
Solution:

Hence, \(\frac{3^{7}}{3^{4} \times 3^{3}}\) = 1

(vi) 2°+ 3°+ 4°
Solution:
1 + 1 + 1 = 3
Hence, 2°+ 3°+ 4° = 3

(vii) 2° × 3° × 4°
Solution:
1 × 1 × 1 = 1
Hence, 2° × 3° × 4° = 1

(viii) (3° + 2°) × 5°
Solution:
(1 + 1) × 1 = 2 × 1 = 2
Hence, (3°+ 2°) × 5° =2

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

(x) \(\left(\frac{a^{5}}{a^{3}}\right)\) × a⁸
Solution:

(xi) \(\frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}}\)
Solution:
a^{5 – 5} × a^{8 – 5} × b^{3 – 2} = 4⁰ × a³ × b
= 1 × a³b = a³b
Hence, \(\frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}}\) = a³b

(xii) (2³ × 2)²
Solution:
(2^{3 + 1})² = (2⁴)² = 2^{4 × 2} = 2⁸
Hence, (2³ × 2)² = 2⁸

Question 3.
Say true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹
Solution:
False: 10 × 10¹¹ = 10¹² and 100¹¹ = 10²²

(ii) 2³ > 5²
Solution:
False: Since 2³ = 8, 5² = 25 and 8 < 25

(iii) 2³ × 3² = 6⁵
Solution:
False: 6⁵ = 2⁵ × 3⁵

(iv) 3° = (1000)°
Solution:
True: 3° = 1 and (1000)°= 1

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
Solution:
(i) 108 × 192
108 = 2² × 3³
192 = 2⁶ × 3

Hence, 108 × 192
= 2² × 3³ × 2⁶ × 3
= 2⁸ × 3⁴

(ii) 270
2 × 3³ × 5

(iii) 729 × 64
729 = 3⁶
64 = 2⁶
Hence,

729 × 64 = 3⁶ × 2⁶
= (3 × 2)⁶ = 6⁶

(iv) 768
2⁸ × 3
Hence,

768 = 2⁸ × 3

Question 5.
Simplify:
(i) \(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}\)
Solution:

(ii) \(\frac{25 \times 5^{2} \times t^{8}}{10^{3} \times t^{4}}\)
25 = 5² ⇒ 10 = 5 × 2
10³ = 5³ × 2³

(iii) \(\frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}}\)
25 = 5² ⇒ 6⁵ = (3 × 2)⁵ = 3⁵ × 2⁵
10⁵ = (5 × 2)⁵ = 5⁵ × 2⁵
Solution:

= 3^{5 – 5} × 5^{5 + 2 – 7} × 2^{5 – 5}
= 3° × 5° × 2°
= 1 × 1 × 1 = 1
Hence, \(\frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}}\) = 1