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If $\log_{10}x-\log_{10}\sqrt x=2 \log_x10$, then a possible value of $x$ is given by

1. $10$
2. $1/100$
3. $1/1000$
4. None of these

Given that, $\log_{10}{x} – \log_{10}{\sqrt{x}} = 2\log_{x}{10}$

$\Rightarrow \log_{10}{x} – \log_{10}{x}^{\frac{1}{2}} = 2\log_{x}{10}$

$\Rightarrow \log_{10}{\left(\frac{x}{x^\frac{1}{2}}\right)} = 2\log_{x}{10} \quad \left[\because \log_{b}{m} – \log_{b}{n} = \log_{b}{\left(\frac{m}{n}\right)} \right]$

$\Rightarrow \log_{10}{\left(x^{1-\frac{1}{2}}\right)} = 2\log_{x}{10}$

$\Rightarrow \log_{10}{x^{\frac{1}{2}}} = 2\log_{x}{10}$

$\Rightarrow \frac{1}{2}\log_{10}{x} = \frac{2}{\log_{x}{10}} \quad \left [\because \log_{b}{a^{k}} = k\log_{b}{a} , \log_{b}{a} = \frac{1}{\log_{a}{b}}\right]$

$\Rightarrow \log_{10}{x} \cdot \log_{10}{x} = 4$

$\Rightarrow \left(\log_{10}{x}\right)^{2} = 4$

$\Rightarrow \left(\log_{10}{x}\right)^{2} = 2^{2}$

$\Rightarrow \log_{10}{x} = 2$

$\Rightarrow x = 10^{2} \quad \left[ \because \log_{b}{x} = a \Rightarrow x=a^{b}\right]$

$\Rightarrow \boxed{x=100}$

Correct Answer $:\text{D}$
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