Question

Which is greater $x={\log }_{3}5ory={\log }_{17}25$?

Answer 1

$x={\log }_{3}5={\log }_{3^2}{5}^2={\log }_{9}25=\frac{1}{{\log }_{25}9}$ using ${\log }_{b}a=\frac{1}{{\log }_{a}b}$

$y={\log }_{17}25=\frac{1}{{\log }_{25}17}$ using ${\log }_{b}a=\frac{1}{{\log }_{a}b}$

We know that $17>9$

${\log }_{25}17>{\log }_{25}9$

$\frac{1}{{\log }_{25}17}<\frac{1}{{\log }_{25}9}$

$\Rightarrow {\log }_{17}25<{\log }_{9}25$

$\Rightarrow {\log }_{17}25<\frac{2}{2}{\log }_{3}5$ using ${\log }_{b^m}{a}^n=\frac{n}{m}\frac{1}{{\log }_{a}b}$

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

$\therefore {\log }_{3}5>{\log }_{17}25$

Hence $x>y$