The sum of the series log_{4} 2 - log_{8} 2 + | vedclass

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(D) Given the series $S = \log_{4} 2 - \log_{8} 2 + \log_{16} 2 - \dots$Using the property $\log_{y^n} x^m = \frac{m}{n} \log_{y} x$,we have $\log_{2^

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$1 - \log_{e} 2$

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(D) Given the series $S = \log_{4} 2 - \log_{8} 2 + \log_{16} 2 - \dots$Using the property $\log_{y^n} x^m = \frac{m}{n} \log_{y} x$,we have $\log_{2^2} 2 = \frac{1}{2} \log_{2} 2 = \frac{1}{2}$,$\log_{2^3} 2 = \frac{1}{3}$,and so on.Thus,$S = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots$We know the logarithmic expansion $\log_{e}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$For $x=1$,$\log_{e}(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$Rearranging this,$\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_{e} 2$.Therefore,$S = 1 - \log_{e} 2$.

The sum of the series $\log_{4} 2 - \log_{8} 2 + \log_{16} 2 - \dots$ is

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