Question

If $0<x<1$, Arrange the following values in the ascending order:

${\log }_{2}x,{\log }_{3}x,{\log }_{e}x,{\log }_{10}x$.

Answer 1

We know that

$2<e<3<10$
Taking log with base $x$ , where $0<x<1$
${\log }_{x}10>{\log }_{x}3>{\log }_{x}e>{\log }_{x}2$

(When base of logarithm is between 0 and 1, and if $m<n\Rightarrow {\log }_{a}m>{\log }_{a}n$ )

$\Rightarrow \frac{1}{{\log }_{10}x}>\frac{1}{{\log }_{3}x}>\frac{1}{{\log }_{e}x}>\frac{1}{{\log }_{2}x}$ ($\because \log a_b=\frac{1}{{\log }_{b}a}$)

$\Rightarrow {\log }_{2}x>{\log }_{e}x>{\log }_{3}x>{\log }_{10}x$
So, in ascending order, ${\log }_{10}x<{\log }_{3}x<{\log }_{e}x<{\log }_{2}x$