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Question

If log0.3(x1)<log0.09(x1), then x lies in the interval


A
(2,)
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B
(2,1)
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C
(1,2)
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D
None
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Solution

The correct option is A (2,)
log0.3 (x1)<log(0.3)2(x1)=12log0.3(x1)
 12log0.3(x1)<0
or log0.3 (x1)<0=log 1
or (x1)>1 or x>2
as base is less than 1, therefore the inequality is reversed, now x>2x lies in (2,)

Mathematics

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