1. Using laws of exponents, simplify and write the answer in exponential form:
(i) 32 × 34 × 38
Solution:
We use the rule of multiply for the power with same base am x an= am+n
= (3) 2+3+8
= 314
(ii) 615 ÷ 610
Solution:
By the rule of dividing for the powers with same base = am ÷ an = am-n
= (6) 15-10
= 65
(iii) a3 × a2
Solution:
= (a) 3+2
= a5
(iv) 7x × 72
Solution:
= (7) x+2
(v) (52) 3 ÷ 53
Solution:
(52) 3 can be written as = (5) 2×3
= 56
Now,
56 ÷ 53
= (5) 6-3
= 53
(vi) 25 × 55
Solution:
= (2 × 5) 5
= 105
(vii) a4 × b4
Solution:
= (a × b) 4
= (ab)4
(viii) (34) 3
Solution:
=(34) 3 can be written as = (3) 4×3
= 312
(ix) (220 ÷ 215) × 23
Solution:
(220 ÷ 215) can be simplified as,
= (220 – 15)
= 25
Then,
25 × 23 can be simplified as,
= (2)5 + 3
= 28
(x) 8t ÷ 82
Solution:
= (8)t – 2
2. Simplify and express each of the following in exponential form:
(i) (23 × 34 × 4)/ (3 × 32)
Solution:
Factors of 4 = 2 × 2
= 22
Factors of 32 = 2 × 2 × 2 × 2 × 2
= 25
Then,
= (23 × 34 × 22)/ (3 × 25)
[we know that, am × an = am+n]
= (25 × 34) / (3 × 25)
= 25-5 × 34-1 … [by: am ÷ an = am-n]
= 20 × 33
= 1 × 33
= 33
(ii) ((52)3 × 54) ÷ 57
Solution:
(52)3 can be written as = (5)2×3
= 56
Now,
= (56 × 54) ÷ 57
= (56+4) ÷ 57
= 510 ÷ 57
= 510-7
= 53
(iii) 254 ÷ 53
Solution:
(25)4 can be written as = (52)4
(52)4 can be written as = (5)2×4
= 58
Then,
= 58 ÷ 53
= 58-3
= 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 × 71 × 115
= 1 × 7 × 115
= 7 × 115
(v) 37/ (34 × 33)
Solution:
= 37/ (34+3)
= 37/ 37
= 37-7
= 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 = (22) 3
(52) 4 can be written as = (2) 2×3
= 26
Now,
= (28 × a5)/ (26 × a3)
= 28-6 × a5-3
= 22 × a2
= 2a2
(x) (a5/a3) × a8
Solution:
= (a5-3) × a8
= a2 × a8
= a10
(xi) (45 × a8b3)/ (45 × a5b2)
Solution:
= 45-5 × (a8-5 × b3-2)
= 40 × (a3b1)
= 1 × a3b
= a3b
(xii) (23 × 2) 2
Solution:
= (23+1) 2
= (24) 2
= 28
3. Say true or false and justify your answer:
(i) 10 × 1011 = 10011
Solution:
According to question:
Let us consider Left Hand Side (LHS) = 10 × 1011
= 101+11
= 1012
Now,
Right Hand Side (RHS) = 10011
= (10 × 10) 11
= (101+1) 11
= (102) 11
= 1022
Now,
By comparing LHS and RHS,
LHS ≠ RHS
Hence, the given statement is false.
(ii) 23 > 52
Solution:
According t question:
Let us consider LHS = 23
23 = 8
Now,
consider RHS = 52
52 = 25
By comparing LHS and RHS,
LHS < RHS
23 < 52
Hence, the given statement is false.
(iii) 23 × 32 = 65
Solution:
According to question:
Let us consider LHS = 23 × 32
23 × 32 = 2 × 2 × 2 × 3 × 3
= 72
Now,
consider RHS = 65
65 = 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.
4. Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
Solution:
According to question:
The factors of 108 = 2 × 2 × 3 × 3 × 3
= 22 × 33
The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
= 26 × 3
Now,
= (22 × 33) × (26 × 3)
= 22+6 × 33+1
= 28 × 34
(ii) 270
Solution:
According to question:
The factors of 270 = 2 × 3 × 3 × 3 × 5
= 2 × 33 × 5
(iii) 729 × 64
Solution:
According to question:
The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2 = 26
Now,
= 36 × 26
(iv) 768
Solution:
According to question:
The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
= 28 × 3
5. Simplify:
(i) ((25)2 × 73)/ (83 × 7)
Solution:
We can write 83 as = (2 × 2 × 2)3
= (23)3
Now we have,
= ((25)2 × 73)/ ((23) 3 × 7)
= (210× 73)/ ((29 × 7)
Now,
= (210-9 × 73-1)
= 2 × 72
= 98
(ii) (25 × 52 × t8)/ (103 × t4)
Solution:
We can write 25 as = 5 × 5 = 52
103 can be written as = (5 × 2) 3
= 53 × 23
Now we have,
= (52 × 52 × t8)/ (53 × 23 × t4)
= (52+2 × t8)/ (53 × 23 × t4)
Now,
= (54 × t8)/ (53 × 23 × t4)
= (54-3 × t8-4)/ 23
= (5t4)/ 8
(iii) (35 × 105 × 25)/ (57 × 65)
Solution:
We can write 105 as = (5 × 2) 5
= 55 × 25
We can write 25 as = 5 × 5 = 52
We can write 65 as = (2 × 3) 5
= 25 × 35
Now we have,
= (35 × 55 × 25 × 52)/ (57 × 25 × 35)
= (35 × 55+2 × 25)/ (57 × 25 × 35)
= (35 × 57 × 25)/ (55 × 25 × 35)
Now,
= (35-5 × 57-7 × 25-5)
= (30 × 50 × 20)
= 1 × 1 × 1
= 1
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