Question

$\frac{1}{{\log }_{3}e}+\frac{1}{{\log }_3{e}^2}+\frac{1}{{\log }_3{e}^4}+....\infty \text{is}$

• A. ${\log }_{e}9$
• B. 0
• C. 1
• D. ${\log }_{e}3$

Answer 1

The correct option is A ${\log }_{e}9$

Solution:

$\frac{1}{{\log }_{3}e}+\frac{1}{{\log }_3{e}^2}+\frac{1}{{\log }_3{e}^4}+.........\infty$

$=\frac{1}{\frac{\log e}{\log 3}}+\frac{1}{\frac{2\log e}{\log 3}}+\frac{1}{4\frac{\log e}{\log 3}}+........\infty$

$=\frac{\log 3}{\log e}[1+\frac{1}{2}+\frac{1}{4}+.......\infty ]$

$={\log }_{e}3[1+\frac{1}{2}+\frac{1}{4}+.......\infty ]$

$={\log }_{e}3\left[\frac{1}{1-\left(\frac{1}{2}\right)}\right]$

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

$={\log }_e{3}^2$

${\log }_{e}9$