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

Given :
\(5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\log_{10}\frac{1}{\sqrt{1-x^2}}\)

Now, we can also express the equation in the following manner :
\(5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\log_{10}(\sqrt{1+x}\times\sqrt{1-x})^{-1}\)

\(5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=(-1)\log_{10}(\sqrt{1+x})+(-1)\log_{10}(\sqrt{1-x})\)

\(5=-\log_{10}\sqrt{1+x}+\log_{10}\sqrt{1+x}-\log_{10}\sqrt{1-x}-4\log_{10}\sqrt{1-x}\)

\(5=-5\log_{10}\sqrt{1-x}\)

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

Now, by squaring on both sides, we get :
\((\sqrt{1-x})^2=\frac{1}{100}\)

∴ \(x=1-\frac{1}{100}=\frac{99}{100}\)

Therefore, \(100x=100\times\frac{99}{100}=99\)
So, the correct answer is 99.