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If $\log_{4} 5 = x$ and $\log_{5} 6 = y$ then $\log_{2} 3$ is equal to

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$2 x y + 1$

$2 x y - 1$

$2 x + 1$

$2 y + 1$

answer is B.

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Detailed Solution

$\log_{4} 5 = x and \log_{5} 6 = y$

$\Rightarrow 5 = 4^{x}$ and $6 = 5^{y}$

$\begin{array}{l} \Rightarrow 6 = {(4^{x})}^{y} \\ \Rightarrow 2 \times 3 = 4^{x y} \\ \Rightarrow 2 \times 3 = {(2^{2})}^{x y} \Rightarrow 3 = 2^{2 x y - 1} \Rightarrow \log_{2} 3 = 2 x y - 1 \end{array}$

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