Question

The sum of the series ${\log }_9\left(3\right)+{\log }_{27}\left(3\right)-{\log }_{81}\left(3\right)+{\log }_{243}\left(3\right)-...$ is

• A. $1-{\log }_e\left(2\right)$
• B. $1+{\log }_e\left(2\right)$
• C. ${\log }_e\left(2\right)$
• D. $1+{\log }_e\left(3\right)$

Answer 1

The correct option is C

${\log }_e\left(2\right)$

Compute the sum:

Given : ${\log }_9\left(3\right)+{\log }_{27}\left(3\right)-{\log }_{81}\left(3\right)+{\log }_{243}\left(3\right)-...$

$$\begin{gathered} {\log }_9\left(3\right)=\frac{\log \left(3\right)}{\log \left(9\right)}\left[\because {\log }_b\left(a\right)=\frac{\log \left(a\right)}{\log \left(b\right)}\right] \\ \Rightarrow {\log }_9\left(3\right)=\frac{\log \left(3\right)}{\log {\left(3\right)}^2}\left[\because \log {\left(a\right)}^b=b\log \left(a\right)\right] \\ \Rightarrow {\log }_9\left(3\right)=\frac{1}{2} \end{gathered}$$

Similarly, ${\log }_{27}\left(3\right)=\frac{1}{3},{\log }_{81}\left(3\right)=\frac{1}{4}.........$

Thus the given series converts into

$$\begin{gathered} \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}........... \\ \Rightarrow \frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}............+\frac{1}{2} \\ \Rightarrow 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}....... \end{gathered}$$

We know that, ${\log }_e\left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.......$

Thus, the series can be written as,

$\Rightarrow {\log }_e\left(2\right)$

Therefore the sum of the series ${\log }_9\left(3\right)+{\log }_{27}\left(3\right)-{\log }_{81}\left(3\right)+{\log }_{243}\left(3\right)-...$ is ${\log }_e\left(2\right)$.

Hence option (C) is the correct option.