Question

Arrange the following values in the ascending order:
${\log }_{2}x,{\log }_{3}x,{\log }_{e}x,{\log }_{10}x$.

If $x>1$

• A. ${\log }_{10}x<{\log }_{3}x<{\log }_{e}x<{\log }_{2}x$
• B. ${\log }_{10}x<{\log }_{e}x<{\log }_{3}x<{\log }_{2}x$
• C. ${\log }_{10}x<{\log }_{3}x<{\log }_{2}x<{\log }_{e}x$
• D. ${\log }_{e}x<{\log }_{10}x<{\log }_{3}x<{\log }_{2}x$

Answer 1

The correct option is A ${\log }_{10}x<{\log }_{3}x<{\log }_{e}x<{\log }_{2}x$

Since $10,3,e,2$ are in decreasing order.
$\Rightarrow {\log }_{x}10>{\log }_{x}3>{\log }_{x}e>{\log }_{x}2[\because x>1]$
$\Rightarrow \frac{1}{{\log }_{x}10}<\frac{1}{{\log }_{x}3}<\frac{1}{{\log }_{x}e}<\frac{1}{{\log }_{x}2}$
$\Rightarrow {\log }_{10}x<{\log }_{3}x<{\log }_{e}x<{\log }_{2}x,[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$