If lo g x ( lo g y k ) > 0 where x , k ∈ ( 0 , 1 ) then y ∈

lo g y k 1 ⇒ k 0 ⇒ k > 1 which is not possible. Case 2: if y y and for lo g y k > 0 ⇒ k < 1 which is true. So y ∈ ( k , 1 ) .

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Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsDomain-RangeMedium2 minQB

MathematicsMediumsingle choice

If $lo g_{x} (lo g_{y} k) > 0$ where $x, k \in (0, 1)$ then $y \in$

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Solution:

$lo g_{y} k < 1$ . Case 1: if $y > 1 \Rightarrow k < y$ ; for $lo g_{y} k > 0 \Rightarrow k > 1$ which is not possible. Case 2: if $y < 1 \Rightarrow k > y$ and for $lo g_{y} k > 0 \Rightarrow k < 1$ which is true. So $y \in (k, 1)$ .

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Domain-Range

⏱ 2mℹ️ Source: QB

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