CAT 2021 Slot 1 QA Question 19: If $5 - \log_{10} \sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \dfrac{1}{\sqrt{1 - x^2}}$, then $100x$ equals

Question

If $5 - \log_{10} \sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \dfrac{1}{\sqrt{1 - x^2}}$, then $100x$ equals

Accepted Answer

Use the facts that $\log \sqrt{a}$ is half of $\log a$ and $1 - x^2 = (1 - x)(1 + x)$.
$5 - \dfrac{1}{2}\log_{10}(1+x) + 2\log_{10}(1-x) = -\dfrac{1}{2}\log_{10}(1-x^2)$
$5 - \dfrac{1}{2}\log_{10}(1+x) + 2\log_{10}(1-x) = -\dfrac{1}{2}\log_{10}(1-x) - \dfrac{1}{2}\log_{10}(1+x)$
The terms containing

If 5−log⁡101+x+4log⁡101−x=log⁡1011−x25 - \log_{10} \sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \dfrac{1}{\sqrt{1 - x^2}}5−log10​1+x​+4log10​1−x​=log10​1−x2​1​, then 100x100x100x equals

If $5 - \log_{10} \sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \dfrac{1}{\sqrt{1 - x^2}}$ , then $100x$ equals