CAT 2021 Slot 1 · Quant · Question 18
CAT 2021 Slot 1 previous year question on Logarithms, with the answer and a step-by-step solution.
If 5-{\log}_{10}\sqrt{1+x}+4{\log}_{10}\sqrt{1-x}={\log}_{10}\frac{1}{\sqrt{1-{x}^{2}}}, then 100x equals
Answer & solution
Answer: 99
Let A={\log}_{10}\sqrt{1+x} and B={\log}_{10}\sqrt{1-x}.
Since \frac{1}{\sqrt{1-{x}^{2}}}=\frac{1}{\sqrt{1+x}\,\sqrt{1-x}}, the right side equals -\left( A+B \right).
The equation becomes 5-A+4B=-A-B, so 5B=-5 and B=-1.
Then \sqrt{1-x}={10}^{-1}, so 1-x=\frac{1}{100} and x=\frac{99}{100}.
Hence 100x=99.
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