Question

If ${\log }_x\left({\log }_{y}k\right)>0$, where $x,k$ $\in (0,1)$, then $y\in$

• A. $(0,\frac{3k}{2})$
• B. $(0,k)$
• C. $(k,1)$
• D. $R^{+}$

Answer 1

The correct option is B $(0,k)$

$lo{g}_x\left(lo{g}_{y}k\right)>0$
$\Rightarrow 0<lo{g}_{y}k<1\because x\in (0,1)$
Case-I : If $y>1\Rightarrow 1<k<y$ which is not possible.
Case-II : If $0<y<1\Rightarrow k<y$ and $k<1$
But $k\in (0,1)\Rightarrow y\in (0,k)$.