A) \[{{\log }_{2}}\alpha ,\,\,{{\log }_{3}}\alpha ,\,\,{{\log }_{e}}\alpha ,\,\,{{\log }_{10}}\alpha \]
B) \[{{\log }_{10}}\alpha ,\,\,{{\log }_{3}}\alpha ,\,\,{{\log }_{e}}\alpha ,\,\,{{\log }_{2}}\alpha \]
C) \[{{\log }_{10}}\alpha ,\,\,{{\log }_{e}}\alpha ,\,\,{{\log }_{2}}\alpha ,\,\,\alpha {{\log }_{3}}\alpha \]
D) None of the above
Correct Answer: B
Solution :
Since,\[10,\,\,3,\,\,e,\,\,2\]are in decreasing order. \[\therefore \]\[{{\log }_{10}}\alpha ,\,\,{{\log }_{3}}\alpha ,\,\,{{\log }_{e}}\alpha ,\,\,{{\log }_{2}}\alpha \]are in increasing order.
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