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
Solution and Explanation
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 \( \boxed{99} \).
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