The given equation is:
\[ 5 - \log_{10} \sqrt{1+x} + 4 \log_{10} \sqrt{1-x} = \log_{10} \frac{1}{\sqrt{1-x^2}} \]
The right-hand side is rewritten as:
\[ 5 - \log_{10} \sqrt{1+x} + 4 \log_{10} \sqrt{1-x} = \log_{10} \left( \sqrt{1+x} \times \sqrt{1-x} \right)^{-1} \]
Using the property \( \log_b(a^n) = n \log_b a \):
\[ 5 - \log_{10} \sqrt{1+x} + 4 \log_{10} \sqrt{1-x} = (-1) \log_{10} \left( \sqrt{1+x} \right) + (-1) \log_{10} \left( \sqrt{1-x} \right) \]
The equation simplifies to:
\[ 5 = -\log_{10} \sqrt{1+x} + \log_{10} \sqrt{1+x} - \log_{10} \sqrt{1-x} - 4 \log_{10} \sqrt{1-x} \]
The equation reduces to:
\[ 5 = -5 \log_{10} \sqrt{1-x} \]
Isolating \( \sqrt{1-x} \):
\[ \sqrt{1-x} = \frac{1}{10} \]
Squaring both sides yields:
\[ (\sqrt{1-x})^2 = \frac{1}{100} \]
This gives:
\[ 1 - x = \frac{1}{100} \]
Therefore:
\[ x = 1 - \frac{1}{100} = \frac{99}{100} \]
The value of \( 100x \) is:
\[ 100x = 100 \times \frac{99}{100} = 99 \]
The final answer is \( \boxed{99} \).
If \[ \log_{p^{1/2}} y \times \log_{y^{1/2}} p = 16, \] then find the value of the given expression.