MCQ
If $y = {2^{1/{{\log }_x}4}}$, then $ x$ is equal to
- A$\sqrt y $
- B$y$
- ✓${y^2}$
- D${y^4}$
$ \Rightarrow {\log _x}4 = \frac{{\log 2}}{{\log y}} $
$\Rightarrow \frac{{{{\log }_e}4}}{{{{\log }_e}x}} = \frac{{{{\log }_e}2}}{{{{\log }_e}y}} $
$\Rightarrow \frac{{2\log 2}}{{\log x}} = \frac{{\log 2}}{{\log y}}$
==> $2\log y = \log x \Rightarrow x = {y^2}$.
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