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$\log _4 2-\log _8 2+\log _{16} 2-\ldots$ is equal to
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Verified Answer
The correct answer is:
$1-\log _e 2$
$\log _4 2-\log _8 2+\log _{16} 2-\ldots$
$=\frac{1}{\log _2 4}-\frac{1}{\log _2 8}+\frac{1}{\log _2 16}-\ldots$
$\left\{\because \log _b a=\frac{1}{\log _a b}\right\}$
$=\frac{1}{\log _2(2)^2}-\frac{1}{\log _2(2)^3}+\frac{1}{\log _2(2)^4}-\ldots$
$=\frac{1}{2 \log _2 2}-\frac{1}{3 \log _2 2}+\frac{1}{4 \log _2 2}-\ldots$ $\left(\because \log _2 2=1\right)$
$=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\ldots\right)$
$\because \log _e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$ (Put $x=1$ )
$\Rightarrow \log _e(1+1)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$
$\Rightarrow \frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\ldots=1-\log _e 2$
$=1-\log _e 2$
$=\frac{1}{\log _2 4}-\frac{1}{\log _2 8}+\frac{1}{\log _2 16}-\ldots$
$\left\{\because \log _b a=\frac{1}{\log _a b}\right\}$
$=\frac{1}{\log _2(2)^2}-\frac{1}{\log _2(2)^3}+\frac{1}{\log _2(2)^4}-\ldots$
$=\frac{1}{2 \log _2 2}-\frac{1}{3 \log _2 2}+\frac{1}{4 \log _2 2}-\ldots$ $\left(\because \log _2 2=1\right)$
$=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\ldots\right)$
$\because \log _e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$ (Put $x=1$ )
$\Rightarrow \log _e(1+1)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$
$\Rightarrow \frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\ldots=1-\log _e 2$
$=1-\log _e 2$
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