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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1
If (4.2)x=(0.42)y=100,{\left( {4.2} \right)^x} = {\left( {0.42} \right)^y} = 100, then 1x−1y=\frac{1}{x} - \frac{1}{y} =
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Correct option: C
12\frac{1}{2}
Option C (1/2) is correct because by taking logarithms and manipulating the equations, we can find that 1/x - 1/y equals 1/2.

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