Question

The value of $\frac{1}{{\log }_{3}e}+\frac{1}{{\log }_3{e}^2}+\frac{1}{{\log }_3{e}^4}+...$ up to infinite terms is

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

Answer 1

Answer: (A)

${\log }_{e}9$

We can write ${\log }_3{e}^2=2{\log }_{3}e$

Similarly, ${\log }_3{e}^4=4{\log }_{3}e$ and so on

Let $t=\frac{1}{{\log }_{3}e}$

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

$=t+\frac{t}{2}+\frac{t}{4}....$

$=t(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....)$

$=t\times \frac{1}{1-\frac{1}{2}}=2t$

Now, $2t=2\frac{1}{{\log }_{3}e}=2\frac{1}{\frac{\log e}{\log 3}}=2\frac{\log 3}{\log e}=2{\log }_{e}3$ $={\log }_{e}9$

Therefore, the answer is ${\log }_{e}9$.