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

Class 7 Maths NCERT Solutions Chapter 13 Exponents and Powers Ex 13.2

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

(i) 3² × 3⁴ × 3⁸
Solution:
3² × 3⁴ × 3⁸ = 3² + 4 + 8

(ii) 6¹⁵ ÷ 6¹⁰
Solution:
6¹⁵ ÷ 6¹⁰ = 6^{15 – 10} = 6⁵

(iii) a³ × a²
Solution:
a³ × a² = a^{3 + 2} = a⁵

(iv) 7^x x 7²
Solution:
7^x × 7² = 7^{x + 2}

(v) (5²)³ ÷ 5³
Solution:
(5²)³ ÷ 5³ = 5^{2 × 3} ÷ 5³
= 5⁶ ÷ 5³ = 5^{6 – 3} = 5³

(vi) 2⁵ × 5⁵
Solution:
2⁵ × 5⁵ = (2 × 5)⁵ = (10)⁵

(vii) a⁴ × b⁴
Solution:
a⁴ × b⁴ = (ab)⁴

(viii) (3⁴)³
Solution:
(3⁴)³ = 3^{4 × 3} = 3¹²

(ix) (2²⁰ ÷ 2¹⁵) × 2³
Solution:
(2²⁰ ÷ 2¹⁵) × 2³ = (2^{10 – 15}) × 2³
= 2⁵ × 2³ = 2^{5 + 3} = 2⁸

(x) 8^t ÷ 8²
Solution:
8^t ÷ 8² = 8^{t – 2}

Question 2.
Simplify and express each of the following in exponential form:
(i) \(\frac{2^{3} \times 3^{4} \times 4}{3 \times 32}\)
Solution:
\(\frac{2^{3} \times 3^{4} \times 4}{3 \times 32}\) = \(\frac{2^{3} \times 3^{4} \times 2^{2}}{3 \times 2^{5}}\)
[∵ 4 = 2 × 2 = 2² , 32 = 2 × 2 × 2 × 2 × 2 = 2⁵]
= \(\frac{2^{3+2} \times 3^{4}}{3^{1} \times 2^{5}}\) = \(\frac{2^{5} \times 3^{4}}{3^{1} \times 2^{5}}\)
= 2^{5 – 5} x 3^{4 – 1} = 2⁰ × 3³ = 1 × 3³ = 3³

(ii) [(5²)³ × 5⁴] ÷ 5⁷
Solution:
[(5²)³ × 5⁴] ÷ 5⁷ = (5 ^{2 × 3} × 5⁴) ÷ 5⁷
=(5⁶ × 5⁴) ÷ 5⁷ = 5^{6 + 4} ÷ 5⁷
= 5¹⁰ ÷ 5⁷ = 5^{10 – 7} = 5³

(iii) 25⁴ ÷ 5³
Solution:
25⁴ ÷ 5³ = (5²)⁴ ÷ 5³
= 5^{2 × 4} ÷ 5³
= 5⁸ ÷ 5³ = 5^{8 – 3}
= 5⁵

(iv) \(\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}\)
Solution:
\(\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}\) = \(\frac{3 \times 7^{2} \times 11^{8}}{3 \times 7 \times 11^{3}}\) = 3^{1 – 1} × 7^{2 – 1} × 11^{8 – 3}
= 3⁰ × 7¹ × 11⁵
= 1 × 7 × 11⁵
= 7 × 11⁵
= 7 × 11⁵

(v) \(\frac{3^{7}}{3^{4} \times 3^{3}}\)
Solution:
\(\frac{3^{7}}{3^{4} \times 3^{3}}\) = \(\frac{3^{7}}{3^{4+3}}\) = \(\frac{3^{7}}{3^{7}}\)
3^{7 – 7} = 3⁰ = 1

(vi) 2⁰ + 3⁰ + 4⁰
Solution:
2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3

(vii) 2⁰ x 3⁰ x 4⁰
Solution:
2⁰ × 3⁰ × 4⁰ = 1 × 1 × 1 = 3

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

(ix) \(\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}\)
Solution:
\(\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}\) = \(\frac{2^{8} \times a^{5}}{\left(2^{2}\right)^{3} \times a^{3}}\) = \(\frac{2^{8} \times a^{5}}{2^{6} \times a^{3}}\)
= 2^{8 – 6} × a^{5 – 3} = 2² × a² = (2a)²

(x) \(\left(\frac{a^{5}}{a^{3}}\right)\) x a⁸
Solution:
\(\left(\frac{a^{5}}{a^{3}}\right)\) × a⁸ = (a^{5 – 3}) × a⁸
= a² × a⁸
= a^{2 + 8}
= a¹⁰

(xi) \(\frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}}\)
Solution:
\(\frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}}\) = 4^{5 – 5} × a^{8 – 5} × b^{3 – 2}
= 4⁰ × a³ × b¹
= 1 × a × b
= a³b

(xii) (2³ × 2)²
Solution:
(2³ × 2)² = (2³ + 1)² = (2⁴)² = 2⁸

Question 3.
Say true or false and justify your answer:

(i) 10 × 10¹¹ = 100¹¹
Solution:

(ii) 2³ > 5²
Solution:

(iii) 2³ × 3² = 6⁵
Solution:

(iv) 3⁰ = (1000)⁰
Solution:

Question 4.
Express each of the following as a product of prime factors only in exponential form:

(i) 108 × 192
Solution:
We have

∴ 108 × 192 = (2 × 2 × 3 × 3 × 3) x (2 × 2 × 2 × 2 × 2 × 2 × 3)
= 2² × 3³ × 2⁶ × 3¹
= 2^{2 + 6} × 3^{3 + 1}
= 2⁸ × 3⁴

(ii) 270
Solution:
We have

∴ 270 = 2 × 3 × 3 ×  3 × 5 = 2 × 3³ × 5

(iii) 729 x 64
Solution:
We have

∴ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3²
64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶
∴ 729 × 64 = 3⁶ × 2⁶

(iv) 768
Solution:
We have

∴ 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3

Question 5.
Simplify:

(i) \(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}\)
Solution:
\(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}\) = \(\frac{2^{10} \times 7^{3}}{\left(2^{3}\right)^{3} \times 7}\) = \(\frac{2^{10} \times 7^{3}}{2^{9} \times 7}\)
= 2^{10 – 9}  × 7^{3 – 1}
= 2¹ × 7²
= 2 × 49
= 98

(ii) \(\frac{\mathbf{2 5} \times \mathbf{5}^{2} \times t^{8}}{10^{3} \times t^{4}}\)
Solution:
\(\frac{25 \times 5^{2} \times t^{8}}{10^{3} \times t^{4}}\) = \(\frac{5^{2} \times 5^{2} \times t^{8}}{(2 \times 5)^{3} \times t^{4}}\) = \(\frac{5^{4} \times t^{8}}{2^{3} \times 5^{3} \times t^{4}}\) = \(\frac{5^{4-3} \times t^{8-4}}{2^{3}}\) = \(\frac{5 \times t^{4}}{2^{3}}\) = \(\frac{5 t^{4}}{8}\)

(iii) \(\frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}}\)
Solution:
\(\frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}}\) = \(\frac{3^{5} \times(2 \times 5)^{5} \times 5^{2}}{5^{7} \times(2 \times 3)^{5}}\) = \(\frac{3^{5} \times 2^{5} \times 5^{5} \times 5^{2}}{5^{7} \times 2^{5} \times 3^{5}}\)
= 3^{5 – 5} × 2^{5 – 5} × 5^{5 + 2 – 7}
= 30 × 20 × 50
= 1 × 1 × 1
= 1