NCERT solution class 7 chapter 13 Exponents and powers 13.2 mathematics

EXERCISE 13.2

---

Q 1 (i, ii, iii, iv), Ex 13.2 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

Q 1 (v, vi, vii, viii, ix, x), Ex 13.2 - Exponents and Powers - Chapter 13 - Maths Class 7th

OPEN IN YOUTUBE

Page No 260:

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) 
(vi) 2⁵ × 5⁵
(vii) a⁴ × b⁴
(viii) (3⁴)³
(ix) 
(x) 8^t ÷ 8²

Answer:

(i) 3² × 3⁴ × 3⁸ = (3)^{2 + 4 + 8} (a^m × aⁿ = a^m+n)
= 3¹⁴
(ii) 6¹⁵ ÷ 6¹⁰ = (6)^{15 − 10} (a^m ÷ aⁿ = a^m−n)
= 6⁵
(iii) a³ × a² = a^{(3 + 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
= 5^{2 × 3} ÷ 5³ (a^m)ⁿ = a^mn
= 5⁶ ÷ 5³
= 5^{(6 − 3)} (a^m ÷ aⁿ = a^m−n)
= 5³
(vi) 2⁵ × 5⁵
= (2 × 5)⁵ [a^m × b^m = (a × b)^m]
= 10⁵
(vii) a⁴ × b⁴
= (ab)⁴ [a^m × b^m = (a × b)^m]
(viii) (3⁴)³ = 3^{4 × 3} = 3¹² (a^m)ⁿ = a^mn
(ix) (2²⁰ ÷ 2¹⁵) × 2³
= (2^{20 − 15}) × 2³ (a^m ÷ aⁿ = a^m−n)
= 2⁵ × 2³
= (2^{5 + 3}) (a^m × aⁿ = a^m+n)
= 2⁸
(x) 8^t ÷ 8² = 8^{(t − 2)} (a^m ÷ aⁿ = a^m−n)

Q 2 (i, ii, iii), Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

Q 2 (iv, v, vi, vii, viii, ix), Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th

OPEN IN YOUTUBE

Q 2 (x, xi, xii), Ex 13.1 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

Question 2:

Simplify and express each of the following in exponential form:
(i)  (ii)  (iii) 
(iv)  (v)  (vi) 2⁰ + 3⁰ + 4⁰
(vii) 2⁰ × 3⁰ × 4⁰ (viii) (3⁰ + 2⁰) × 5⁰ (ix) 
(x)  (xi)  (xii) 

Answer:

(i)

(ii) [(5²)³ × 5⁴] ÷ 5⁷
= [5^{2 × 3} × 5⁴] ÷ 5⁷ (a^m)ⁿ = a^mn
= [5⁶ × 5⁴] ÷ 5⁷
= [5^{6 + 4}] ÷ 5⁷ (a^m × aⁿ = a^m+n)
= 5¹⁰ ÷ 5⁷
= 5^{10 − 7} (a^m ÷ aⁿ = a^m−n)
= 5³
(iii) 25⁴ ÷ 5³ = (5 ×5)⁴ ÷ 5³
= (5²)⁴ ÷ 5³
= 5^{2 × 4} ÷ 5³ (a^m)ⁿ = a^mn
= 5⁸ ÷ 5³
= 5^{8 − 3} (a^m ÷ aⁿ = a^m−n)
= 5⁵
(iv)

= 1 × 7 × 11⁵ = 7 × 11⁵
(v)

(vi) 2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3
(vii) 2⁰ × 3⁰ × 4⁰ = 1 × 1 × 1 = 1
(viii) (3⁰ + 2⁰) × 5⁰ = (1 + 1) × 1 = 2
(ix)

(x)

(xi)

(xii) (2³ × 2)² =  (a^m × aⁿ = a^m+n)
= (2⁴)² = 2^{4 × 2} (a^m)ⁿ = a^mn
= 2⁸

Q 3, Ex 13.2 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

Question 3:

Say true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹ (ii) 2³ > 5²
(iii) 2³ × 3² = 6⁵ (iv) 3⁰ = (1000)⁰

Answer:

(i) 10 × 10¹¹ = 100¹¹
L.H.S. = 10 × 10¹¹ = 10^{11 + 1} (a^m × aⁿ = a^m+n)
= 10¹²
R.H.S. = 100¹¹ = (10 ×10)¹¹= (10²)¹¹
= 10^{2 × 11} = 10²² (a^m)ⁿ = a^mn
As L.H.S. ≠ R.H.S.,
Therefore, the given statement is false.
(ii) 2³ > 5²
L.H.S. = 2³ = 2 × 2 × 2 = 8
R.H.S. = 5² = 5 × 5 = 25
As 25 > 8,
Therefore, the given statement is false.
(iii) 2³ × 3² = 6⁵
L.H.S. = 2³ × 3² = 2 × 2 × 2 × 3 × 3 = 72
R.H.S. = 6⁵ = 7776
As L.H.S. ≠ R.H.S.,
Therefore, the given statement is false.
(iv) 3⁰ = (1000)⁰
L.H.S. = 3⁰ = 1
R.H.S. = (1000)⁰ = 1 = L.H.S.
Therefore, the given statement is true.

Q 4, Ex 13.2 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

Page No 261:

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 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)
= (2² × 3³) × (2⁶ × 3)
= 2^{6 + 2} × 3^{3 + 1} (a^m × aⁿ = a^m+n)
= 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

Q 5, Ex 13.2 - Exponents and Powers - Chapter 13 - Maths Class 7th - NCERT

OPEN IN YOUTUBE

Question 5:

Simplify:
(i)  (ii)  (iii) 

Answer:

(i)

(ii)

(iii)