If 5- \log_10 \sqrt{1+\x} + 4\log_10 \sqrt{1-\x} = \log_10 1/\sqrt{1-\x^2} , then 100x equals

CAT Exam 2021 Slot 1 Solved Paper

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5- \log_10 √{1+\x} + 4\log_10 √{1-\x} = \log_10 1/√{1-\x^2}

, then 100x equals

Solution:

5- \log_10 √{1+\x} + 4\log_10 √{1-\x} = \log_10 1/√{1-\x^2}

we can re-write the equition as :

5- \log_10 √{1+\x} \ 4 \log_10 √{1-\x} = \log_10 (√{1+\x} × √{1-\x} )^{-1}

5-\log_10 √{1+\x} + 4\log_10 √{1-\x} =(-1) \log_10(√{1+\x}) + (-1) \log_10 (√{1-\x})

5= -\log_10 √{1+\x} + \log_10 √{1+\x} -\log_10 √{1-\x} -4\log_10 √{1-\x}

5= -5\log_10 √{1-\x}

√{1-\x} = 1/100

squaring both sides :

(√{1-\x})^2 =1/100

∴

\x =1- 1/100 =99/100

hence,

100\x = 100× 99/100 =99

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