Question

Arrange in ascending order:
${\log }_{2}x,{\log }_{3}x,{\log }_{e}x,{\log }_{4}x$ in ascending order, if
(a) $x>1$
(b) $0<x<1$

Answer 1

$\because 2<e<3<10$

(i) For $x>1$ we have

${\log }_{x}2<{\log }_{x}e<{\log }_{x}3<{\log }_{x}10$

By Base change theorem,

$\Rightarrow \frac{1}{{\log }_{2}x}<\frac{1}{{\log }_{e}x}<\frac{1}{{\log }_{3}x}<\frac{1}{{\log }_{10}x}$

By the Basic inequality if $\frac{1}{a}<\frac{1}{b}$ then $a>b$

$\Rightarrow {\log }_{2}x>{\log }_{e}x>{\log }_{3}x>{\log }_{10}x$

Hence, ascending order is

${\log }_{10}x<{\log }_{3}x<{\log }_{e}x<{\log }_{2}x$

(ii) For $0<x<1$ we have

${\log }_{x}2>{\log }_{x}e>{\log }_{x}3>{\log }_{x}10$

By base change theorem,

$\Rightarrow \frac{1}{{\log }_{2}x}>\frac{1}{{\log }_{e}x}>\frac{1}{{\log }_{3}x}>\frac{1}{{\log }_{10}x}$

By the basic inequality,if $\frac{1}{a}>\frac{1}{b}$ then $a<b$

Thus, ${\log }_{2}x<{\log }_{e}x<{\log }_{3}x<{\log }_{10}x$ which is in ascending order.