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Question

If log0.3(x1)<log0.09(x1), then x lies in the interval 
  1. (2,)
  2. (1,2)
  3. (2,1)
  4. None of these


Solution

The correct option is A (2,)
log0.3(x1)<log0.09(x1)log0.3(x1)<log0.32(x1)(x1)2>(x1)(x1)(x2)>0x(,1)(2,)
Now from log definition
x1>0x>1x(1,)
x(2,)

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