NCERT Class 7 Mathematics

(Ron) #1
EXPONENTS AND POWERS 259

(ii) 2^3 × 2^2 × 5^5 = 23+2 × 5^5
=2^5 × 55 = (2 × 5)^5 = 10^5
(iii) 66 624 3  = 6624 3

=


(^6633)
3
(^666)
6



(iv)  
2663

 235 
 =[2

(^6) × 3 (^6) ]× 56
= 23 5^66
= 
235 ^6
=^306
(v) 8 = 2 × 2 × 2 = 2^3
Therefore 8^2 ÷ 2^3 =(2^3 )^2 ÷ 2^3
=2^6 ÷ 2^3 = 26–3 = 23
EXAMPLE 12 Simplify :
(i)
43
32
1294
6827

 (ii) 2
(^3) ×a (^3) × 5a (^4) (iii) 23 2
94
45
2
SOLUTION
(i) We have
43
32
1294
6827




24232
3323
(2 ×3) ×(3 ) ×2
(2×3) ×(2 ) ×3


2332
232 3
2233
2233
2 4 4 23 2
33233
8246
3633
()() × =
×


82 46
36 33
2×3
2×3


10 10
96
2×3
2×3
=210 – 9 × 310 – 6 = 2^1 × 3^4
= 2 × 81 = 162
(ii) 2^3 × a^3 × 5a^4 = 2^3 ×a^3 × 5 × a^4
=2^3 × 5 × a^3 × a^4 = 8 × 5 × a3 + 4
= 40 a^7

Free download pdf