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

These NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 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 ÷ 5³
(vi) 2⁵ × 5⁵
(vii) a⁴ × b⁴
(viii) (3⁴)³
(ix) (2²⁰ ÷ 2¹⁵) × 2³
(x) 8^t ÷ 8²
Answer:
(i) 3² × 3⁴ × 3⁸ = 3^{2 + 4 + 8}
(a^m × aⁿ × ^r = a^m+n+r) = 3¹⁴
(ii) 6¹⁵ ÷ 6¹⁰ = \(\frac{6^{15}}{6^{10}}\) = 6^15-10
(\(\frac{a^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n) = 6⁵
(iii) a³ × a² = a³⁺²
(a^m × aⁿ = a^m+n ) = a⁵
(iv) 7^x × 7² = 7^x+2 (a^m × aⁿ = a^m+n )
(v) (5²)³ ÷ 5³

(vi) 2⁵ × 5⁵ = (2 × 5)⁵
[a^m × b^m = (ab)^m]
= (ab)⁴
(vii) a⁴ x b⁴
(viii) (3⁴)³ = 3¹²
[(a^m)ⁿ = a^mn]
(ix) (2²⁰ ÷ 2¹⁵) x 2³ = (\(\frac{2^{20}}{2^{15}}\)) x 2³
= (2^20-15) × 2³ [ \(\frac{a^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n]
= 2⁵ × 2³
= 2^{5 + 3} (a^m x aⁿ = a^m+n) = 2⁸
(x) 8^t ÷ 8² = \(\frac{8^{t}}{8^{2}}=8^{t-2}\left(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}=\mathrm{a}^{\mathrm{m}-\mathrm{n}}\right)\)

Question 2.
Simplify and express each of the following in exponential form:
(i) \(\)
(ii) [(5²)³ × 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° × 3° × 4°
(viii) (3° + 2°) × 5°
(ix) \(\frac{2^{8} \times \mathrm{a}^{5}}{4^{3} \times \mathrm{a}^{3}}\)
(x) \(\left(\frac{a^{5}}{a^{3}}\right) \times a^{8}\)
(xi) \(\frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}}\)
(xii) (2³ × 2)²
Answer:

= 2° × 3³
= 1 × 3³ = 3³
(a° = 1)

(iv) \(\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}=\frac{3 \times 7^{2} \times 11^{8}}{3 \times 7 \times 11^{3}}\)
= 3^1-1 × 7^2-1 × 11^8-3
(\(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n)
= 3° × 7¹ × 11⁵
= 1 × 7 × 11⁵
= 7 ×11⁵

(vi) 2° + 3°+ 4° = 1 + 1 + 1 = 3 (a° = 1)
(vii) 2° x 3° × 4° = 1 × 1 × 1 = 3 (a° = 1)
(viii) (3° + 2°) × 5° = (1 + 1) × 1 = 3 (a° = 1)
= 2 × 1 = 2

(ix) \(\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}=\frac{2^{8} \times a^{5}}{\left(2^{2}\right)^{3} \times a^{3}}\)
= \(\frac{2^{8} \times a^{5}}{2^{6} \times a^{3}}\) [(a^m)ⁿ = a^mn]
= 2^8-6 × a ^5-3 (\(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n)
= 2² × a²
= 2a²

(x) \(\left(\frac{a^{5}}{a^{3}}\right)\) x a⁸ = (a^5-3) x a⁸
[(a^m)ⁿ = a^mn]
= a² × a⁸ (a^m x aⁿ = a^m+n)
= a²⁺⁸
= a¹⁰

(xi) \(\frac{4^{5} a^{8} b^{3}}{4^{5} a^{5} b^{2}}=\frac{4^{5-5} \times a^{8} \times b^{3}}{a^{5} \times b^{2}}\)
= 4° × a^8-5 × b^3-2 ( \(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n)
= 1 × a³ × b
= a³b (a⁰ = 1)

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

Question 3.
Say true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹
(ii) 2³ > 5²
(iii) 2³ × 3² = 6⁵
(iv) 3° = (1000)°
Answer:
(i) 10 × 10¹¹ = 100¹¹
L.H.S = 10 x 10¹¹
= 10⁽¹⁺¹¹⁾= 10¹²
(a^m × aⁿ = a^m+n)
R.H.S = 100¹¹
= (10²)H= 10²²
(a^m)ⁿ = a^mn
L.H.S ≠ R.H.S.
∴ 10 × 10¹¹ ≠ 100¹¹
This statement is false.

(ii) 2³ > 5²
2³ = 2 × 2 × 2 = 8
5² = 5 × 5 = 25
8 < 25
i. e. 2³ > 5²
This statement is false.

(iii) 2³ × 3² = 6⁵
L.H.S = 2³ × 3²
= 2 × 2 × 2 × 3 × 3 = 72
R.H.S = 6⁵ = 6 × 6 × 6 × 6 × 6
= 36 × 36 × 6
= 7776
L.H.S ≠ R.H.S.
72 ≠ 7776
∴ 2³ × 3² ≠ 6⁵
This statement is false.

(iv) 3° = (1000)°
L.H.S = 3° = 1
R.H.S = (1000)°= 1
3° = (1000)°
L.H.S = R.H.S
3° = (1000)°
∴ This statement is true.

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 x 2 x 3 x 3 x 3 x 2 x 2

= 2⁸ × 3⁴

(ii) 270 = 2 × 3 × 3 × 3 × 5

= 2 × 3³ × 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) \(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}\)
(ii) \(\frac{25 \times 5^{2} \times t^{8}}{10^{3} \times t^{4}}\)
(iii) \(\frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}}\)
Answer:
(i) \(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}=\frac{2^{10} \times 7^{3}}{\left(2^{3}\right)^{3} \times 7}\)
[(a^m)ⁿ = a^mn ]
= \(\frac{2^{10} \times 7^{3}}{2^{9} \times 7}\)
= 2^10-9 × 7^3-1 [\(\frac{a^{m}}{a^{n}}\) = a^m-n]
= 2¹ × 7²
= 2 × 49 = 98
Thus, \(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}\) = 98

= 3^5-5 × 2^5-5 × 5^7-7
= 3° × 2° × 5°
= 1 × 1 × 1 = 1