If x = log_{0.1} 0.001 and y = log_9 81, then | 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 $y = \log_9 81$. Sinc

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$\sqrt{2} - 1$

Vedclass · Answer

(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, $\sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1$.

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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