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If $y$ is a negative number such that $2^{y^{2}\log _{2}3}=5^{\log_{2}3}$, then $y$ equals

1. $\log _{2}\left ( \frac{1}{3} \right )$
2. $-\log _{2}\left ( \frac{1}{3} \right )$
3. $\log _{2}\left ( \frac{1}{5} \right )$
4. $-\log _{2}\left ( \frac{1}{5} \right )$

Given that, $2^{y^{2} \log_{3}5} = 5^{\log_{2}3}; y < 0$

Taking $\log_{3}$ on both sides.

$\log_{3} 2^{y^{2} \log_{3}5} = \log_{3} 5^{\log_{2}3}$

$\Rightarrow y^{2} \log_{3}5 \; \log_{3}2 = \log_{2}3 \; \log_{3}5$

$\Rightarrow y^{2} \log_{3}2 = \log_{2}3$

$\Rightarrow y^{2} = \dfrac{\log_{2}3}{\log_{3}2}$

$\Rightarrow y^{2} = \dfrac{\log_{2}3}{\frac{1}{\log_{2}3}} \quad \left[ \because \log_{a}b = \dfrac{1}{\log_{b}a}\right]$

$\Rightarrow y^{2} = \log_{2}3 \cdot \log_{2}3$

$\Rightarrow y^{2} = \left( \log_{2}3 \right)^{2}$

$\Rightarrow y = \; – \log_{2}3 \quad [\because y \;\text{is negative}]$

$\Rightarrow y = \log_{2}3^{-1}$

$\Rightarrow \boxed{ y = \log_{2} \frac{1}{3}}$

Correct Answer$: \text{A}$
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