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If $5 – \log_{10} \sqrt{1+x} + 4 \log_{10} \sqrt{1-x} = \log_{10} \frac{1}{\sqrt{1-x^{2}}},$ then $100x$ equals
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Given that,

$5 – \log_{10} \sqrt{1+x} + 4 \log_{10} \sqrt{1-x} = \log_{10} \frac{1}{\sqrt{1-x^{2}}}$

$ \Rightarrow 5 – \log_{10} (\sqrt{1+x}) + 4 \log_{10} (\sqrt{1-x}) = \log_{10} \left( 1-x^{2} \right)^{\frac{-1}{2}} $

$ \Rightarrow 5 – \log_{10} (\sqrt{1+x}) + 4 \log_{10} (\sqrt{1-x}) =\; – \log_{10} \left( \sqrt{1-x^{2}} \right) \quad [\because \log_{b}a^{-x} =\; – \log_{b}a^{x}]$

$ \Rightarrow 5 – \log_{10} (\sqrt{1+x}) + 4 \log_{10} (\sqrt{1-x}) =\; – \log_{10} \left( (\sqrt{1-x})(\sqrt{1+x}) \right)$

$ \Rightarrow 5 – \log_{10} (\sqrt{1+x}) + 4 \log_{10} (\sqrt{1-x}) =\; – \log_{10} (\sqrt{1-x}) – \log_{10} (\sqrt{1+x}) \quad [\because \log_{b}m + \log_{b}n = \log_{b}(mn)]$

$ \Rightarrow 5 \log_{10} (\sqrt{1-x}) = \;– 5$

$ \Rightarrow \log_{10} (\sqrt{1-x}) =\; – 1$

$ \Rightarrow \sqrt{1-x} = 10^{– 1} \quad [\because \log_{b}a = x \Rightarrow a = b^{x}]$

$ \Rightarrow \sqrt{1-x} = \frac{1}{10}$

Squares on both sides

$ \Rightarrow 1 – x = \frac{1}{100}$

$ \Rightarrow x = 1 – \frac{1}{100}$

$ \Rightarrow \boxed{ x = \frac{99}{100}}$

$ \therefore$ The value of $100 x = 100 \times \frac{99}{100} = 99.$

Correct Answer $: 99$
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