Question

If \(5 - log_{10}\  \sqrt {1 + x }+ 4\  log_{10 }\  \sqrt {1-x} = log_{10}\  \frac {1}{\sqrt {1-x^2}}\), then \(100 x \) equals

Answer 1

We are given the equation:

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

Step 1: Simplifying the equation

Expressing the right-hand side in a different form: \[ 5 - \log_{10} \sqrt{1+x} + 4 \log_{10} \sqrt{1-x} = \log_{10} \left( \sqrt{1+x} \times \sqrt{1-x} \right)^{-1} \]

Step 2: Further simplifying

Applying the logarithmic property \( \log_b(a^n) = n \log_b a \), we get: \[ 5 - \log_{10} \sqrt{1+x} + 4 \log_{10} \sqrt{1-x} = (-1) \log_{10} \left( \sqrt{1+x} \right) + (-1) \log_{10} \left( \sqrt{1-x} \right) \]

Step 3: Grouping terms

This simplifies to: \[ 5 = -\log_{10} \sqrt{1+x} + \log_{10} \sqrt{1+x} - \log_{10} \sqrt{1-x} - 4 \log_{10} \sqrt{1-x} \]

Step 4: Further simplification

The equation simplifies to: \[ 5 = -5 \log_{10} \sqrt{1-x} \]

Step 5: Solving for \( x \)

Solving for \( \sqrt{1-x} \): \[ \sqrt{1-x} = \frac{1}{10} \] Squaring both sides: \[ (\sqrt{1-x})^2 = \frac{1}{100} \] Thus: \[ 1 - x = \frac{1}{100} \] So: \[ x = 1 - \frac{1}{100} = \frac{99}{100} \]

Step 6: Final Answer

The value of \( 100x \) is: \[ 100x = 100 \times \frac{99}{100} = 99 \]

Conclusion:

The correct answer is \( \boxed{99} \).