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

NCERT Solutions Class 7 Mathematics
Chapter – 13 (Exponents and Powers)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 13 Exponents and Powers  Exercise 13.2 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 13 Exponents and Powers

• NCERT Solution Class 7 Maths Exercise – 13.1
• NCERT Solution Class 7 Maths Exercise – 13.3

Exercise – 13.2

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

Solution –

(i) 3² × 3⁴ × 3⁸
By the rule of multiplying the powers with same base = a^m × aⁿ = a^{m + n}
Then,
= (3)^{2 + 4 + 8}
= 3¹⁴

(ii) 6¹⁵ ÷ 6¹⁰
By the rule of dividing the powers with same base = a^m ÷ aⁿ = a^{m – n}
Then,
= (6)^{15 – 10}
= 6⁵

(iii) a³ × a²
By the rule of multiplying the powers with same base = a^m × aⁿ = a^{m + n}
Then,
= (a)^{3 + 2}
= a⁵

(iv) 7^x × 7²
By the rule of multiplying the powers with same base = a^m × aⁿ = a^{m + n}
Then,
= (7)^{x + 2}

(v) (5²)³ ÷ 5³
By the rule of taking power of as power = (a^m)ⁿ = a^mn
(5²)³ can be written as = (5)^{2 × 3}
= 5⁶
Now, 5⁶ ÷ 5³
By the rule of dividing the powers with same base = a^m ÷ aⁿ = a^{m – n}
Then,
= (5)^{6 – 3}
= 5³

(vi) 2⁵ × 5⁵
By the rule of multiplying the powers with same exponents = a^m × b^m = ab^m
Then,
= (2 × 5)⁵
= 10⁵

(vii) a⁴ × b⁴
By the rule of multiplying the powers with same exponents = a^m × b^m = ab^m
Then,
= (a × b)⁴
= ab⁴

(viii) (3⁴)³
By the rule of taking power of as power = (a^m)ⁿ = a^mn
(3⁴)³ can be written as = (3)^{4 × 3}
= 3¹²

(ix) (2²⁰ ÷ 2¹⁵) × 2³
By the rule of dividing the powers with same base = a^m ÷ aⁿ = a^{m – n}
(2²⁰ ÷ 2¹⁵) can be simplified as,
= (2)^{20 – 15}
= 2⁵
Then,
By the rule of multiplying the powers with same base = a^m × aⁿ = a^{m + n}
2⁵ × 2³ can be simplified as,
= (2)^{5 + 3}
= 2⁸

(x) 8^t ÷ 8²
By the rule of dividing the powers with same base = a^m ÷ aⁿ = a^{m – n}
Then,
= (8)^{t – 2}

2. Simplify and express each of the following in exponential form:
(i)

(ii) ((5²)³ × 5⁴) ÷ 5⁷

(iii) 25⁴ ÷ 5³

(iv)

(v)

(vi) 2⁰ + 3⁰ + 4⁰

(vii) 2⁰ × 3⁰ × 4⁰

(viii) (3⁰ + 2⁰) × 5⁰

(ix)

(x)

(xi)

(xii) (2³ × 2)²

Solution –

(i)
Factors of 32 = 2 × 2 × 2 × 2 × 2
= 2⁵
Factors of 4 = 2 × 2
= 2²
Then,
=

=     [∵ a^m × aⁿ = a^{m + n}]

= 2^{5 – 5} × 3^{4 – 1} [∵a^m ÷ aⁿ = a^{m – n}]
= 2⁰ × 3³
= 1 × 3³
= 3³

(ii) ((5²)³ × 5⁴) ÷ 5⁷
(5²)³ can be written as = (5)^{2 × 3}  [∵(a^m)ⁿ = a^mn]
= 5⁶
Then,
= (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
(25)⁴ can be written as = (5 × 5)⁴
= (5²)⁴
(5²)⁴ can be written as = (5)^{2 × 4}  [∵(a^m)ⁿ = a^mn]
= 5⁸
Then,
= 5⁸ ÷ 5³
= 5^{8 – 3} [∵a^m ÷ aⁿ = a^{m – n}]
= 5⁵

(iv)

Factors of 21 = 7 × 3
Then,
=

= 3^1-1 × 7^2-1 × 11^{8 – 3}
= 3⁰ × 7 × 11⁵
= 1 × 7 × 11⁵
= 7 × 11⁵

(v)

= [∵a^m × aⁿ = a^{m + n}]

=
= 3^{7 – 7} [∵a^m ÷ aⁿ = a^{m – n}]
= 3⁰
= 1

(vi) 2⁰ + 3⁰ + 4⁰
= 1 + 1 + 1
= 3

(vii) 2⁰ × 3⁰ × 4⁰
= 1 × 1 × 1
= 1

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

(ix)
(4)³ can be written as = (2 × 2)³
= (2²)³
(2²)³ can be written as = (2)^{2 × 3}   [∵(a^m)ⁿ = a^mn]
= 2⁶
Then,
=

= 2^{8 – 6} × a^{5 – 3}  [∵a^m ÷ aⁿ = a^{m – n}]
= 2² × a²
= 2a² [∵(a^m)ⁿ = a^mn]

(x)

= (a^{5 – 3}) × a⁸  [∵a^m ÷ aⁿ = a^{m – n}]
= a² × a⁸
= a^{2 + 8}  [∵a^m × aⁿ = a^{m + n}]
= a¹⁰

(xi)

= 4^{5 – 5} × (a^{8 – 5} × b^{3 – 2})  [∵a^m ÷ aⁿ = a^{m – n}]
= 4⁰ × (a³b)
= 1 × a³b
= a³b

(xii) (2³ × 2)²
= (2^{3 + 1})²  [∵a^m × aⁿ = a^{m + n}]
= (2⁴)²
(2⁴)² can be written as = (2)^{4 × 2}  [∵(a^m)ⁿ = a^mn]
= 2⁸

3. Say true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹
(ii) 2³ > 5²
(iii) 2³ × 3² = 6⁵
(iv) 3²⁰ = (1000)⁰

Solution –

(i) 10 × 10¹¹ = 100¹¹
LHS = 10 × 10¹¹
= 10¹⁺¹¹
= 10¹²
RHS = 100¹¹
= (10²)¹¹
= 10²²
10¹² ≠ 10²²
LHS ≠ RHS
Hence, the given statement is false.

(ii) 2³ > 5²
LHS = 2³ = 8
RHS = 5²2 = 25
8 < 25
∴ 2³ < 5²
LHS < RHS
2³ < 5²
Hence, the given statement is false.

(iii) 2³ × 3² = 6⁵
LHS = 2³ × 3²
= 8 × 9
= 72
RHS = 6⁵
= 6 × 6 × 6 × 6 × 6
= 7776
∴ 72 ≠ 7776
LHS ≠ RHS
Hence, the given statement is false.

(iv) 3⁰ = (1000)⁰
⇒ 1 = 1 True [∵ a⁰ = 1]
LHS = RHS
3⁰ = 1000⁰
Hence, the given statement is true.

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

The factors of 108 = 2 × 2 × 3 × 3 × 3
= 2² × 3³
The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
= 2⁶ × 3
= (2² × 3³) × (2⁶ × 3)
= 2^{2 + 6} × 3^{3 + 1}  [∵a^m × aⁿ = a^{m + n}]
= 2⁸ × 3⁴

(ii) 270

The factors of 270 = 2 × 3 × 3 × 3 × 5
= 2 × 3³ × 5

(iii) 729 × 64

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3
= 3⁶
The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2
= 2⁶
Then,
= (3⁶ × 2⁶)
= 3⁶ × 2⁶

(iv) 768

The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
= 2⁸ × 3

5. Simplify:
(i)

(ii)

(iii)

Solution –

(i)

8³ can be written as = (2 × 2 × 2)³
= (2³)³
We have,

=

=   [∵(a^m)ⁿ = a^mn]

= (2^{10 – 9} × 7^{3 – 1}) [∵a^m ÷ aⁿ = a^{m – n}]
= 2 × 7²
= 2 × 7 × 7
= 98

(ii)

25 can be written as = 5 × 5
= 5²
10³ can be written as = 10³
= (5 × 2)³
= 5³ × 2³
We have,

=

= [∵a^m × aⁿ = a^{m + n}]

= [∵a^m ÷ aⁿ = a^{m – n}]

=

=

(iii)

10⁵ can be written as = (5 × 2)⁵
= 5⁵ × 2⁵
25 can be written as = 5 × 5
= 5²
6⁵ can be written as = (2 × 3)⁵
= 2⁵ × 3⁵

=

= [∵a^m × aⁿ = a^{m + n}]

= (3^{5 – 5} × 5^{7 – 7} × 2^{5 – 5})
= (3⁰ × 5⁰ × 2⁰)  [∵a^m ÷ aⁿ = a^{m – n}]
= 1 × 1 × 1
= 1