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.