Logarithm problem : Prove that $lo{g}_{{3}^{2}}\frac{1}{2}>0$

My approach :

$lo{g}_{{3}^{2}}\frac{1}{2}>0$

$\Rightarrow \frac{1}{2}lo{g}_{3}\frac{1}{2}>0$

$\Rightarrow \frac{1}{2}[lo{g}_{3}1-lo{g}_{3}2]>0$

$\Rightarrow \frac{1}{2}[0-lo{g}_{3}2]>0$

$\Rightarrow -\frac{1}{2}[lo{g}_{3}2]>0$ { which is false}

Please suggest... thanks...

My approach :

$lo{g}_{{3}^{2}}\frac{1}{2}>0$

$\Rightarrow \frac{1}{2}lo{g}_{3}\frac{1}{2}>0$

$\Rightarrow \frac{1}{2}[lo{g}_{3}1-lo{g}_{3}2]>0$

$\Rightarrow \frac{1}{2}[0-lo{g}_{3}2]>0$

$\Rightarrow -\frac{1}{2}[lo{g}_{3}2]>0$ { which is false}

Please suggest... thanks...