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Question

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

A
x(0,)
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B
x(1,)
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C
x(2,)
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D
x(3,)
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Solution

The correct option is C x(2,)
log0.3(x1)<log0.09(x1)
log0.3(x1)12log0.3(x1)<0
log0.3(x1)<0
x1>1, as 0< base <1
x>2

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