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# CAT 2021 Quant Question [Slot 1] with Solution 16

Question
If $5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = {\log _{10}}\frac{1}{{\sqrt {1 - {x^2}} }}$, then 100x equals
Option: 99
Solution:
>$5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = {\log _{10}}\frac{1}{{\sqrt {1 - {x^2}} }}$
We can re-write the equation as: $5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = {\log _{10}}{\left( {\sqrt {1 + x} \times \;\sqrt {1 - x} } \right)^{ - 1}}$
$5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = \left( { - 1} \right){\log _{10}}\left( {\sqrt {1 + x} } \right) + \left( { - 1} \right){\log _{10}}\left( {\sqrt {1 - x} } \right)$
$5 = - {\log _{10}}\sqrt {1 + x} + {\log _{10}}\sqrt {1 + x} - {\log _{10}}\sqrt {1 - x} - 4{\log _{10}}\sqrt {1 - x}$
$5 = - 5{\log _{10}}\sqrt {1 - x}$
$\sqrt {1 - x} = \frac{1}{{10}}$
Squaring both sides: ${\left( {\sqrt {1 - x} } \right)^2} = \frac{1}{{100}}$
$\therefore \;$ $x = 1 - \frac{1}{{100}} = \frac{{99}}{{100}}$
Hence, $100\;x\; = 100 \times \;\frac{{99}}{{100}} = 99$
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