Question

Show that ${\log }_{3}5<{\log }_{17}25$

Answer 1

Since, $3<\sqrt{1}7$
Taking logarithm with base $5$
${\log }_{5}3<{\log }_5\sqrt{17}$

${\log }_{5}3<\frac{1}{2}{\log }_{5}17$ ($\because \log x^m=m\log x$)

$\Rightarrow {\log }_{5}3<{\log }_{5^2}17$ $(\because {\log }_{a^b}x=\frac{1}{b}{\log }_{a}x)$

$\Rightarrow {\log }_{5}3<{\log }_{25}17$

Hence, ${\log }_{25}17$ is greater.