## EX 1.6 QUESTION 1.

Find:

(i)641/2

(ii)321/5

(iii)1251/3

Solution:

(i)641/2

641/2

= (8×8)1/2

= (82)½

= 81 [⸪2×1/2 = 2/2 =1]

= 8

(ii)321/5

321/5

= (25)1/5

= (25)

= 21 [⸪5×1/5 = 1]

= 2

(iii)1251/3

(125)1/3

= (5×5×5)1/3

= (53)

= 51 (3×1/3 = 3/3 = 1)

= 5

## EX 1.6 QUESTION 2.

Find:

(i) 93/2

(ii) 322/5

(iii)163/4

(iv) 125-1/3

Solution:

(i) 93/2

93/2

= (3×3)3/2

= (32)3/2

= 33 [⸪2×3/2 = 3]

=27

(ii) 322/5

322/5 = (2×2×2×2×2)2/5

= (25)2⁄5

= 22 [⸪5×2/5= 2]

= 4

(iii)163/4

163/4

= (2×2×2×2)3/4

= (24)3⁄4

= 23 [⸪4×3/4 = 3]

= 8

(iv) 125-1/3

125-1/3

= (5×5×5)-1/3

= (53)-1⁄3

= 5-1 [⸪3×-1/3 = -1]

= 1/5

## EX 1.6 QUESTION 3.

Simplify:

(i) 22/3×21/5

(ii) (1/33)7

(iii) 111/2/111/4

(iv) 71/2×81/

Solution:

(i) 22/3×21/5

22/3×21/5

= 2(2/3)+(1/5)                   [ am×an=am+n]

= 213/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]

(ii) (1/33)7

(1/33)= (3-3)7                [(am)= am x n]

= 3-21

(iii) 111/2/111/4

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

71/2×81/2 = (7×8)1/2             [am×b= (a×b)m ]

= 561/2

Categories: class 9NCERT NOTES

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