Question

${\log }_{e}3–{\log }_e\frac{9}{2^2}+{\log }_e\frac{27}{3^2}–{\log }_e\frac{81}{4^2}+\dots$ is

• A. ${\log }_{e}3{\log }_{e}2$
• B. ${\log }_{e}3$
• C. ${\log }_{e}2$
• D. $\frac{{\log }_{e}5}{{\log }_{e}3}$

Answer 1

The correct option is A

${\log }_{e}3{\log }_{e}2$

Explanation for the correct option:

Finding ${\mathbf{log}}_{\mathbf{e}}\mathbf{3}\mathbf{}\mathbf{–}\mathbf{}{\mathbf{log}}_{\mathbf{e}}\mathbf{}\frac{\mathbf{9}}{{\mathbf{2}}^{\mathbf{2}}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{log}}_{\mathbf{e}}\mathbf{}\frac{\mathbf{27}}{{\mathbf{3}}^{\mathbf{2}}}\mathbf{}\mathbf{–}\mathbf{}{\mathbf{log}}_{\mathbf{e}}\mathbf{}\frac{\mathbf{81}}{{\mathbf{4}}^{\mathbf{2}}}\mathbf{}\mathbf{+}\mathbf{\dots }$is

Given: ${\log }_{e}3–{\log }_e\frac{9}{2^2}+{\log }_e\frac{27}{3^2}–{\log }_e\frac{81}{4^2}+\dots$

$$\begin{gathered} ={\log }_{e}3–{\log }_e\frac{9}{2^2}+{\log }_e\frac{27}{3^2}–{\log }_e\frac{81}{4^2}+\dots \\ ={\log }_{e}3–\frac{{\log }_e{3}^2}{2^2}+\frac{{\log }_e{3}^3}{3^2}–\frac{{\log }_e{3}^4}{4^2}+\dots \\ ={\log }_{e}3–\frac{(2{\log }_{e}3)}{2^2}+\frac{(3{\log }_{e}3)}{3^2}–\frac{(4{\log }_{e}3)}{4^2}+.. \\ ={\log }_{e}3–\frac{({\log }_{e}3)}{2}+\frac{({\log }_{e}3)}{3}–\frac{({\log }_{e}3)}{4}+.. \\ ={\log }_{e}3(1–\frac{1}{2}+\frac{1}{3}–\frac{1}{4}\dots .)\mathbf{\left[}\mathbf{\because }\mathbf{ln}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\right]} \\ =({\log }_{e}3)({\log }_{e}2) \end{gathered}$$

Hence the correct answer is option (A).