If x=log _{0.1} 0.001 and y=log _9 81,then sq | vedclass

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(C) Given $x = \log_{0.1} 0.001$. Since $0.001 = (0.1)^3$,we have $x = \log_{0.1} (0.1)^3 = 3 \log_{0.1} 0.1 = 3(1) = 3$. Given $y = \log_9 81$. Since

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$\sqrt{2}-1$

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(C) Given $x = \log_{0.1} 0.001$. Since $0.001 = (0.1)^3$,we have $x = \log_{0.1} (0.1)^3 = 3 \log_{0.1} 0.1 = 3(1) = 3$. Given $y = \log_9 81$. Since $81 = 9^2$,we have $y = \log_9 9^2 = 2 \log_9 9 = 2(1) = 2$. Now,we need to evaluate $\sqrt{x - 2\sqrt{y}}$. Substituting the values of $x$ and $y$,we get $\sqrt{3 - 2\sqrt{2}}$. We can write $3 - 2\sqrt{2}$ as $(\sqrt{2})^2 + 1^2 - 2(\sqrt{2})(1) = (\sqrt{2} - 1)^2$. Therefore,$\sqrt{x - 2\sqrt{y}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1$.

If $x=\log _{0.1} 0.001$ and $y=\log _9 81$,then $\sqrt{x-2 \sqrt{y}}$ is equal to

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