Question

answer of all plz
with full explanation...​

Answer 1

Answer:

tell me which book and exercise is it

(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

Answer 2

now you have to do that 2 to the power 0 + 3 to the power 0 + 4 to the power 0 equals to 1 + 1 + 1 = 3