log _4 2 - log _8 2 + log _{16} 2 - dots is e | vedclass

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(D) The given series is $\log _4 2 - \log _8 2 + \log _{16} 2 - \dots$Using the property $\log _b a = \frac{1}{\log _a b}$, we get:$= \frac{1}{\log _2

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$1 - \log _e 2$

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(D) The given series is $\log _4 2 - \log _8 2 + \log _{16} 2 - \dots$Using the property $\log _b a = \frac{1}{\log _a b}$, we get:$= \frac{1}{\log _2 4} - \frac{1}{\log _2 8} + \frac{1}{\log _2 16} - \dots$$= \frac{1}{\log _2(2^2)} - \frac{1}{\log _2(2^3)} + \frac{1}{\log _2(2^4)} - \dots$$= \frac{1}{2 \log _2 2} - \frac{1}{3 \log _2 2} + \frac{1}{4 \log _2 2} - \dots$Since $\log _2 2 = 1$, the expression becomes:$= \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots$We know the 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 the terms, we get $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log _e 2$.

$\log _4 2 - \log _8 2 + \log _{16} 2 - \dots$ is equal to

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