Question

${\log }_{4}2-{\log }_{8}2+{\log }_{16}2+...$is

• A. $e^2$
• B. ${\log }_{e}2+1$
• C. ${\log }_{e}3-2$
• D. $1-{\log }_{e}2$

Answer 1

The correct option is D

$1-{\log }_{e}2$

Explanation for correct answer:

Step-1 : Simplifying each term:

Let $S={\log }_{4}2-{\log }_{8}2+{\log }_{16}2+...$

We know that ${\log }_{a}b=\frac{1}{{\log }_{b}a}$

$$\begin{gathered} {\log }_{4}2=\frac{1}{{\log }_{2}4} \\ =\frac{1}{{\log }_2\left(2\times 2\right)} \\ =\frac{1}{{\log }_{2}2+{\log }_{2}2}\left[\because lo{g}_a\left(b\times c\right)=lo{g}_{a}b+lo{g}_{a}c\right] \\ =\frac{1}{1+1}\mathbf{}\left[\because lo{g}_{a}a=1\right] \\ =\frac{1}{2} \end{gathered}$$

$$\begin{gathered} {\log }_{8}2=\frac{1}{{\log }_{2}8} \\ =\frac{1}{{\log }_2\left(2\times 2\times 2\right)} \\ =\frac{1}{{\log }_{2}2+{\log }_{2}2+{\log }_{2}2}\left[\because lo{g}_a\left(b\times c\right)=lo{g}_{a}b+lo{g}_{a}c\right] \\ =\frac{1}{1+1+1}\left[\because lo{g}_{a}a=1\right] \\ =\frac{1}{3} \end{gathered}$$

$$\begin{gathered} {\log }_{16}2=\frac{1}{{\log }_{2}16} \\ =\frac{1}{{\log }_2\left(2\times 2\times 2\times 2\right)} \\ =\frac{1}{{\log }_{2}2+{\log }_{2}2+{\log }_{2}2+{\log }_{2}2}\mathbf{[}\mathbf{\because }\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{a}}\mathbf{}\left(b\times c\right)\mathbf{=}\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{a}}\mathbf{}\mathbf{b}\mathbf{}\mathbf{+}\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{a}}\mathbf{}\mathbf{c}\mathbf{}\mathbf{]} \\ =\frac{1}{1+1+1+1}\mathbf{[}\mathbf{\because }\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{a}}\mathbf{}\mathbf{a}\mathbf{=}\mathbf{1}\mathbf{]} \\ =\frac{1}{4} \end{gathered}$$

Step-2 : Sum of series :

Therefore,$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-...\to \left(i\right)$

We know that $\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{}\mathbf{=}\mathit{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}$

Substitute $x=1$,

$$\begin{gathered} \log (1+1)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+... \\ \log 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+... \end{gathered}$$

Simplifying equation$\left(i\right)$

$$\begin{gathered} S=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-... \\ =1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-...\mathbf{[}\mathbf{\text{Add and subtract by}}\mathbf{}\mathbf{1}\mathbf{]} \\ =1-\left(1-\frac{1}{2}+\frac{1}{3}-...\right) \\ =1-\log 2\mathbf{[}\mathbf{\because }\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{2}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{-}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{]} \end{gathered}$$

Hence, the correct answer is option (D).