$\frac{(l o g x - l o g y) (l o g x^{2} + l o g y^{2})}{(l o g x^{2} - l o g y^{2}) (l o g x + l o g y)}$ is equal to

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l o g \frac{x}{y}

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log xy

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Solution

The correct option is B 1
We have,
$= \frac{(l o g x - l o g y) (l o g x^{2} + l o g y^{2})}{(l o g x^{2} - l o g y^{2}) (l o g x + l o g y)}$
We know that
$l o g m^{n} = n l o g m$
Therefore,
$= \frac{(l o g x - l o g y) (2 l o g x + 2 l o g y)}{(2 l o g x - 2 l o g y) (l o g x + l o g y)}$

$= \frac{2 (l o g x - l o g y) (l o g x + l o g y)}{2 (l o g x - l o g y) (l o g x + l o g y)}$

$= 1$
Hence, this is the answer.

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