Sequence And Series Question 71
Question: The value of $ {\log_3}e-{\log_9}e+{\log_{27}}e…. $ is equal to
Options:
A) $ {\log_3}2 $
B) $ {\log_2}3 $
C) $ 2{\log_3}2 $ v
D) None of the se
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Answer:
Correct Answer: A
Solution:
$ {\log_3}e-{\log_9}e+{\log_{27}}e-…. $ $ =\frac{1}{{\log_{e}}3}[ 1-\frac{1}{2}+\frac{1}{3}-….\infty ] $ $ =\frac{{\log_{e}}2}{{\log_{e}}3}={\log_3}2 $ .
BETA
