A) \[{{\log }_{3}}2\]
B) \[{{\log }_{2}}3\]
C) \[2{{\log }_{3}}2\] v
D) None of the se
Correct Answer: A
Solution :
\[{{\log }_{3}}e-{{\log }_{9}}e+{{\log }_{27}}e-....\]\[=\frac{1}{{{\log }_{e}}3}\left[ 1-\frac{1}{2}+\frac{1}{3}-....\infty \right]\] \[=\frac{{{\log }_{e}}2}{{{\log }_{e}}3}={{\log }_{3}}2\].
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