Question

If $x={\log }_{3}5$ and $y={\log }_{17}25$, then show that $x>y$

Answer 1

Given $y={\log }_{17}25$

$\frac{1}{y}={\log }_{25}17=\frac{1}{2}{\log }_{5}17[\because {\log }_{a^m}b=\frac{1}{m}{\log }_{a}b]$

Also given $x={\log }_{3}5$

$\frac{1}{x}={\log }_{5}3=\frac{1}{2}{\log }_{5}9[\because m\log b=\log b^m]$

Since, $17>9$

${\log }_{5}17>{\log }_{5}9$

$\Rightarrow \frac{1}{2}{\log }_{5}17>\frac{1}{2}{\log }_{5}9$

$\therefore \frac{1}{y}>\frac{1}{x}$

$\Rightarrow x>y$