If (4.2)^{x} = (0.42)^{y} = 100,then frac{1}{ | vedclass

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(C) Given $(4.2)^{x} = 100$ and $(0.42)^{y} = 100$.From $(4.2)^{x} = 100$,we have $4.2 = 100^{1/x} = (10^2)^{1/x} = 10^{2/x}$.From $(0.42)^{y} = 100$,

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$1/2$

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(C) Given $(4.2)^{x} = 100$ and $(0.42)^{y} = 100$.From $(4.2)^{x} = 100$,we have $4.2 = 100^{1/x} = (10^2)^{1/x} = 10^{2/x}$.From $(0.42)^{y} = 100$,we have $0.42 = 100^{1/y} = (10^2)^{1/y} = 10^{2/y}$.We know that $4.2 / 0.42 = 10$.Substituting the values,we get $10^{2/x} / 10^{2/y} = 10^1$.Using the laws of exponents,$10^{(2/x - 2/y)} = 10^1$.Equating the exponents,$2/x - 2/y = 1$.Dividing by $2$,we get $\frac{1}{x} - \frac{1}{y} = \frac{1}{2}$.

If $(4.2)^{x} = (0.42)^{y} = 100$,then $\frac{1}{x} - \frac{1}{y} = $

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