Class IX Math NCERT Solutions For Real Numbers Exercise– 1.6

Exercise– 1.6
1.  Find:
       (i)   641/2
       (ii)   321/5
       (iii)   1251/3
Ans.
       (i)   ⇒ 64 = 8 × 8 = 82
              ⇒ (64)1/2 = (82)1/2 = 82 × 1/2
                    [⇒ (am)n = am× n]
              ⇒ 81 = 8
              Thus, 641/2 = 8
       (ii)   ⇒ 32 = 2×2×2×2×2× = 25
              ⇒ (32)1/5 = (25)1/5 = 25×1/5
              [⇒(am)n = amn]
              = 21 = 2
              Thus, (32)1/5 = 2
       (iii)   ⇒ 125 = 5×5×5 = 53
              ⇒ (125)1/3 = (53)1/3 = 53×1/3 = 51
              Thus, 1251/3 = 5
2.  Find:
       (i)   93/2
       (ii)   322/5
       (iii)   163/4
       (iv)   125-1/3
Ans.
       (i)   ⇒ 9 = 3×3 = 32
              ⇒ (9)3/2 (32)3/2 = 32×3/2 = 33 = 27
              Thus, 93/2 = 27
       (ii)   ⇒ 32 = 2×2×2×2×2× = 25
              ⇒ (32)2/5 = (25)2/5 = 22 = 4
              Thus, 322/5 = 4
       (iii)   ⇒ 16 = 2×2×2×2 = 24
              ⇒ (16)3/4 = (24)3/4 = 24×3/4 = 23 = 8
              Thus, 163/4 = 8
       (iv)   ⇒ 125 = 5×5×5 = 53
              (125)-1/3 (53)-1/3
3.  Simplify:
     
Ans.
     
     
       (iv)   ⇒ am · bm = (ab)m
                ⇒ 71/2 · 81/2 = (7 × 8)1/2 = (56)1/2
                Thus, 71/2 · 81/2 = (56)1/2

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