NCERT Solutions Class 7 Math Chapter 11 Exponents and Powers Exercise 11.2

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

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

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

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

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

(vi) 2⁵ × 5⁵
Solution: By the rule of multiplying the powers with same exponents = am × bm = abm
Then,
= (2 ×5)⁵
= 10⁵

(vii) a⁴ × b⁴
Solution: By the rule of multiplying the powers with same exponents = am × bm = abm
Then,
= (a × b)⁴
= ab⁴

(viii) (3⁴)³
Solution: By the rule of taking power of as power = (am)n = amn
(34)3 can be written as = (3)^{4 × 3}
= 3¹²

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

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

(i) (23 × 34 × 4)/ (3 × 32)
Solution: Factors of 32 = 2 × 2 × 2 × 2 × 2
= 25
Factors of 4 = 2 × 2
= 22
Then,
= (23 × 34 × 22)/ (3 × 25)
= (23 + 2 × 34) / (3 × 25) … [∵am × an = am + n]
= (25 × 34) / (3 × 25)
= 25 – 5 × 34 – 1 … [∵am ÷ an = am – n]
= 20 × 33
= 1 × 33
= 33

(ii) ((52)3 × 54) ÷ 57
Solution: (52)3 can be written as = (5)2 × 3 … [∵(am)n = amn]
= 56
Then,
= (56 × 54) ÷ 57
= (56 + 4) ÷ 57 … [∵am × an = am + n]
= 510 ÷ 57
= 510 – 7 … [∵am ÷ an = am – n]
= 53

(iii) 254 ÷ 53
Solution: (25)4 can be written as = (5 × 5)4
= (52)4
(52)4 can be written as = (5)2 × 4 … [∵(am)n = amn]
= 58
Then,
= 58 ÷ 53
= 58 – 3 … [∵am ÷ an = am – n]
= 55

(iv) (3 × 72 × 118)/ (21 × 113)
Solution: Factors of 21 = 7 × 3
Then,
= (3 × 72 × 118)/ (7 × 3 × 113)
= 31-1 × 72-1 × 118 – 3
= 30 × 7 × 115
= 1 × 7 × 115
= 7 × 115

(v) 37/ (34 × 33)
Solution:  37/ (34+3) … [∵am × an = am + n]
= 37/ 37
= 37 – 7 … [∵am ÷ an = am – n]
= 30
= 1

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

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

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

(ix) (28 × a5)/ (43 × a3)
Solution: (4)3 can be written as = (2 × 2)3
= (22)3
(52)4 can be written as = (2)2 × 3 … [∵(am)n = amn]
= 26
Then,
= (28 × a5)/ (26 × a3)
= 28 – 6 × a5 – 3 … [∵am ÷ an = am – n]
= 22 × a2
= 2a2 … [∵(am)n = amn]

(x) (a5/a3) × a8
Solution: (a5 – 3) × a8 … [∵am ÷ an = am – n]
= a2 × a8
= a2 + 8 … [∵am × an = am + n]
= a10

(xi) (45 × a8b3)/ (45 × a5b2)
Solution:
= 45 – 5 × (a8 – 5 × b3 – 2) … [∵am ÷ an = am – n]
= 40 × (a3b)
= 1 × a3b
= a3b

(xii) (23 × 2)2
Solution:
= (23 + 1)2 … [∵am × an = am + n]
= (24)2
(24)2 can be written as = (2)4 × 2 … [∵(am)n = amn]
= 28

(i) 10 × 1011 = 10011
Solution: Let us consider Left Hand Side (LHS) = 10 × 1011
= 101 + 11 … [∵am × an = am + n]
= 1012
Now, consider Right Hand Side (RHS) = 10011
= (10 × 10)11
= (101 + 1)11
= (102)11
= (10)2 × 11 … [∵(am)n = amn]
= 1022
By comparing LHS and RHS,
LHS ≠ RHS
Hence, the given statement is false.

(ii) 23 > 52
Solution: Let us consider LHS = 23
Expansion of 23 = 2 × 2 × 2
= 8
Now, consider RHS = 52
Expansion of 52 = 5 × 5
= 25
By comparing LHS and RHS,
LHS < RHS
23 < 52
Hence, the given statement is false.

(iii) 23 × 32 = 65
Solution: Let us consider LHS = 23 × 32
Expansion of 23 × 32= 2 × 2 × 2 × 3 × 3
= 72
Now, consider RHS = 65
Expansion of 65 = 6 × 6 × 6 × 6 × 6
= 7776
By comparing LHS and RHS,
72 ≠ 7776
LHS ≠ RHS
Hence, the given statement is false.

(iv) 30 = (1000)0
Solution: Let us consider LHS = 30
= 1
Now, consider RHS = 10000
= 1
By comparing LHS and RHS,
LHS = RHS
30 = 10000
Hence, the given statement is true.

(i) 108 × 192
Solution: The factors of 108 = 2 × 2 × 3 × 3 × 3
= 22 × 33
The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
= 26 × 3
Then,
= (22 × 33) × (26 × 3)
= 22 + 6 × 33 + 1 … [∵am × an = am + n]
= 28 × 34

(ii) 270
Solution: The factors of 270 = 2 × 3 × 3 × 3 × 5
= 2 × 33 × 5

(iii) 729 × 64
Solution:  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
Solution: The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
= 28 × 3

(i) ((25)2 × 73)/ (83 × 7)
Solution: 83 can be written as = (2 × 2 × 2)3
= (23)3
We have,
= ((25)2 × 73)/ ((23)3 × 7)
= (25 × 2 × 73)/ ((23 × 3 × 7) … [∵(am)n = amn]
= (210 × 73)/ (29 × 7)
= (210 – 9 × 73 – 1) … [∵am ÷ an = am – n]
= 2 × 72
= 2 × 7 × 7
= 98

(ii) (25 × 52 × t8)/ (103 × t4)
Solution: 25 can be written as = 5 × 5
= 52
103 can be written as = 103
= (5 × 2)3
= 53 × 23
We have,
= (52 × 52 × t8)/ (53 × 23 × t4)
= (52 + 2 × t8)/ (53 × 23 × t4) … [∵am × an = am + n]
= (54 × t8)/ (53 × 23 × t4)
= (54 – 3 × t8 – 4)/ 23 … [∵am ÷ an = am – n]
= (5 × t4)/ (2 × 2 × 2)
= (5t4)/ 8

(iii) (35 × 105 × 25)/ (57 × 65)
Solution: 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
Then we have,
= (35 × 55 × 25 × 52)/ (57 × 25 × 35)
= (35 × 55 + 2 × 25)/ (57 × 25 × 35) … [∵am × an = am + n]
= (35 × 57 × 25)/ (57 × 25 × 35)
= (35 – 5 × 57 – 7 × 25 – 5)
= (30 × 50 × 20) … [∵am ÷ an = am – n]
= 1 × 1 × 1
= 1