If $5-\log _{10} \sqrt{1+x}+4 \log _{10} \sqrt{1-x}=\log _{10} \frac{1}{\sqrt{1-x^{2}}}$,then $100 \mathrm{x}$ equals

We need to use properties of logarithms to simplify and solve this logarithmic equation.

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The equation starts with the number 5. To work with logarithms effectively, we'll convert this to logarithmic form:

$5 = \log_{10}(10^5) = \log_{10}(100000)$

By definition of logarithms, if $\log_{10}(a) = b$, then $10^b = a$. Since $10^5 = 100000$, we have $\log_{10}(100000) = 5$.

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Now our equation becomes:

$\log_{10}(100000) - \log_{10}\sqrt{1+x} + 4\log_{10}\sqrt{1-x} = \log_{10}\frac{1}{\sqrt{1-x^2}}$

Using these key logarithm properties:

$a\log b = \log(b^a)$

$\log a + \log b = \log(ab)$

$\log a - \log b = \log(\frac{a}{b})$

$4\log_{10}\sqrt{1-x} = \log_{10}(\sqrt{1-x})^4$

Combining all terms on the left side:

$\log_{10}\left[\frac{100000 \times (\sqrt{1-x})^4}{\sqrt{1+x}}\right] = \log_{10}\frac{1}{\sqrt{1-x^2}}$

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Since both sides have the same logarithm base, and the logarithms are equal, their arguments must be equal:

$\frac{100000 \times (\sqrt{1-x})^4}{\sqrt{1+x}} = \frac{1}{\sqrt{1-x^2}}$

Key insight: When $\log_a(P) = \log_a(Q)$, then $P = Q$.

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Notice that $\sqrt{1-x^2} = \sqrt{(1+x)(1-x)} = \sqrt{1+x} \cdot \sqrt{1-x}$

So our equation becomes:

$\frac{100000 \times (\sqrt{1-x})^4}{\sqrt{1+x}} = \frac{1}{\sqrt{1+x} \cdot \sqrt{1-x}}$

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$100000 \times (\sqrt{1-x})^4 = \frac{1}{\sqrt{1-x}}$

$100000 \times (\sqrt{1-x})^4 \times \sqrt{1-x} = 1$

$100000 \times (\sqrt{1-x})^5 = 1$

Therefore: $(\sqrt{1-x})^5 = \frac{1}{100000} = 10^{-5}$

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Taking the fifth root:

$\sqrt{1-x} = (10^{-5})^{\frac{1}{5}} = 10^{-1} = \frac{1}{10}$

$1-x = \left(\frac{1}{10}\right)^2 = \frac{1}{100}$

Therefore: $x = 1 - \frac{1}{100} = \frac{99}{100}$

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$100x = 100 \times \frac{99}{100} = 99$