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Logarithmic Equation

CAT 2021 Slot 2 · Quantitative Ability · Hard · Logarithms

This is a hard Quantitative Ability question from the CAT 2021 Slot 2 paper. It tests Logarithms. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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If 5−log⁡101+x+4log⁡101−x=log⁡1011−x25 - \log_{10}\sqrt{1 + x} + 4\log_{10}\sqrt{1 - x} = \log_{10}\frac{1}{\sqrt{1 - x^2}}, then 100x100x equals

Answer

99

Explanation

Simplify the log terms: 5−12log⁡10(1+x)+2log⁡10(1−x)=−12log⁡10(1−x2)5 - \frac{1}{2}\log_{10}(1+x) + 2\log_{10}(1-x) = -\frac{1}{2}\log_{10}(1-x^2) Notice that log⁡10(1−x2)=log⁡10(1+x)+log⁡10(1−x)\log_{10}(1-x^2) = \log_{10}(1+x) + \log_{10}(1-x). So, 5−12log⁡10(1+x)+2log⁡10(1−x)=−12log⁡10(1+x)−12log⁡10(1−x)5 - \frac{1}{2}\log_{10}(1+x) + 2\log_{10}(1-x) = -\frac{1}{2}\log_{10}(1+x) - \frac{1}{2}\log_{10}(1-x)

Cancel −12log⁡10(1+x)-\frac{1}{2}\log_{10}(1+x) from both sides: 5+2log⁡10(1−x)=−12log⁡10(1−x)5 + 2\log_{10}(1-x) = -\frac{1}{2}\log_{10}(1-x) 5=−52log⁡10(1−x)5 = -\frac{5}{2}\log_{10}(1-x) log⁡10(1−x)=−2\log_{10}(1-x) = -2 1−x=10−2=0.011 - x = 10^{-2} = 0.01 x=0.99x = 0.99 Therefore, 100x=99100x = 99.

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