Today i am solving the ncert class 9th chapter number system . How i solve the exercise 1.6 question number 3 . Give me the best and easiest solution of this question also find the best method of this question.Simplify: (i) 22/3×21/5,(ii) (1/33)7,(iii) 111/2/111/4,(iv) 71/2×81 .
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Simplify: (i) 22/3×21/5,(ii) (1/33)7,(iii) 111/2/111/4,(iv) 71/2×81 .Q.3(i),(ii),(iii),(iv)
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(i) 22/3×21/5
Solution:
22/3×21/5 = 2(2/3)+(1/5) [⸪Since, am×an=am+n____ Laws of exponents]
= 213/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]
(ii) (1/33)7
Solution:
(1/33)7 = (3-3)7 [⸪Since,(am)n = am x n____ Laws of exponents]
= 3-21
(iii) 111/2/111/4
Solution:
111/2/111/4 = 11(1/2)-(1/4)
= 111/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]
(iv) 71/2×81/2
Solution:
71/2×81/2 = (7×8)1/2 [⸪Since, (am×bm = (a×b)m ____ Laws of exponents]
= 561/2