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Question

If logx(logyk)>0 where x,k(0,1) then y

A
(0, x)
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B
(0, k)
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C
(k, 1)
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D
R+
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Solution

The correct option is C (k, 1)
logyk<1
case 1 : if y > 1 k < y
for logyk>0k>1 which is not possible
case 2 : if y < 1 k > y
and for logyk>0k<1 which is true

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