If frac{1}{2} le log_{0.1} x le 2,then which | vedclass

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(B) Given the inequality: $\frac{1}{2} \le \log_{0.1} x \le 2$. Since the base $0.1$ is between $0$ and $1$,the inequality sign reverses when we remov

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The value of $x$ lies between $\frac{1}{100}$ and $\frac{1}{\sqrt{10}}$.

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(B) Given the inequality: $\frac{1}{2} \le \log_{0.1} x \le 2$. Since the base $0.1$ is between $0$ and $1$,the inequality sign reverses when we remove the logarithm: $(0.1)^2 \le x \le (0.1)^{1/2}$. Calculating the values: $(0.1)^2 = (\frac{1}{10})^2 = \frac{1}{100}$. $(0.1)^{1/2} = \sqrt{0.1} = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}$. Thus,the range for $x$ is $\frac{1}{100} \le x \le \frac{1}{\sqrt{10}}$. This means $x$ lies between $\frac{1}{100}$ and $\frac{1}{\sqrt{10}}$.

If $\frac{1}{2} \le \log_{0.1} x \le 2$,then which of the following is true?

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