Question 4.

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

We can re-write the equation as:

$5 - \log_{10}\sqrt{1+x} + 4\log_{10}\sqrt{1-x} = \log_{10}\left(\sqrt{1+x}\times\sqrt{1-x}\right)^{-1}$

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

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

$5 = -5\log_{10}\sqrt{1-x}$

$\sqrt{1-x} = \frac{1}{10}$

Squaring both sides:

$(\sqrt{1-x})^2 = \frac{1}{100}$

$\therefore\ x = 1 - \frac{1}{100} = \frac{99}{100}$

Hence, $100x = 100 \times \frac{99}{100}$ = 99