Question

The value of the expression $\frac{(\log x-\log y)(\log x^2+\log y^2)}{(\log x^2-\log y^2)(\log x+\log y)}$ is equal to

• A. $0$
• B. $1$
• C. $\log \frac{x}{y}$
• D. $\log xy$

Answer 1

The correct option is B $1$

Let $E=\frac{(\log x-\log y)(\log x^2+\log y^2)}{(\log x^2-\log y^2)(\log x+\log y)}$

Using the formula,

$\log a^x=x\log a$

Therefore, $E=\frac{(\log x-\log y)(2\log x+2\log y)}{(2\log x-2\log y)(\log x+\log y)}$

$\Rightarrow E=\frac{(\log x-\log y)\times 2\times (\log x+\log y)}{2\times (\log x-\log y)\times (\log x+\log y)}$

$\Rightarrow E=1$

Hence, option B is correct.