Q100 / 160 · TS EAMCET 2010 · Mathematics

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

  • A
    $e^2$
  • B
    $\log _e 2$
  • C
    $1+\log _e 3$
  • Correct answer
    D
    $1-\log _e 2$

Solution

$\log _4 2-\log _8 2+\log _{16} 2-\ldots$ $=\frac{1}{\log _2 4}-\frac{1}{\log _2 8}+\frac{1}{\log _2 16}-\ldots$ $\left\{\because \log _b a=\frac{1}{\log _a b}\right\}$ $=\frac{1}{\log _2(2)^2}-\frac{1}{\log _2(2)^3}+\frac{1}{\log _2(2)^4}-\ldots$ $=\frac{1}{2 \log _2 2}-\frac{1}{3 \log _2 2}+\frac{1}{4 \log _2 2}-\ldots$ $\left(\because \log _2 2=1\right)$ $=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\ldots\right)$ $\because \log _e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$ (Put $x=1$ ) $\Rightarrow \log _e(1+1)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$ $\Rightarrow \frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\ldots=1-\log _e 2$ $=1-\log _e 2$

Hence, the correct answer is option: 'D'.

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