Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev

Class 9 Mathematics by Full Circle

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Class 9 : Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev

The document Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.
All you need of Class 9 at this link: Class 9

Ques 1: Find: 
(i) 641/2 
(ii) 321/5 
(iii) 1251/3
Ans: (i) ∵ 64 = 8 x 8 = 82
∴ (64)1/2 = (82)1/2 = 82x1/2   [∵ (am)n = ax n]
= 81 = 8
Thus, 641/2 = 8
(ii) ∵ 32 = 2 x 2 x 2 x 2 x 2 = 25
∴ (32)1/5 = (25)1/5 = 25x1/5   [∵ (am)n = amn]
= 21 = 2
Thus, (32)1/5 = 2
(iii) ∵ 125 = 5 x 5 x 5 = 53

∴ (125)1/3 = (53)1/3 = 53x1/3 = 51

Thus, 1251/3 = 5

Ques 2: Find:
(i) 93/2 
(ii) 322/5 
(iii) 163/4 
(iv) 125–1/3
Ans: (i) ∵ 9 = 3 x 3 = 32
∴ (9)3/2 (32)3/2 = 32 x 3/2 = 33 = 27
Thus, 93/2 = 27
(ii) ∵ 32 = 2 x 2 x 2 x 2 x 2 = 25
∴ (32)2/5 = (25)2/5 = 25x2/5 = 22 = 4

Thus, 322/5 = 4
(iii) ∵ 16 = 2 x 2 x 2 x 2 = 24
∴ (16)3/4 = (24)3/4 = 24x3/4 = 23 = 8
Thus, 163/4 = 8
(iv) ∵ 125 = 5 x 5 x 5 = 53
∴ (125)–1/3 = (53)–1/3 = 53x(–1/3) = 5–1 = 1/5

Ques 3: Simplify:

(i) 22/3 . 21/5

Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev
Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev
(iv) 71/2 . 81/2

Ans: (i) ∵ am . an = am+n
∴ 22/3 . 21/5 = 22/3 + 1/5
= 213/15   Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev
(ii) ∵ Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev

Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev

Thus, 

Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev

(iii)
Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev  Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev

Thus, 

Ex 1.6 NCERT Solutions - Number System Class 9 Notes | EduRev

(iv) ∵ am . bm = (ab)m

∴ 71/2 . 81/2 = (7 x 8)1/2 = (56)1/2

Thus, 71/2 . 81/2 = (56)1/2

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