<s>[INST] Logarithm problem : Prove that $log_{3^2} \frac{1}{2} > 0$ Logarithm problem : 
Prove that $log_{3^2} \frac{1}{2} > 0$
My approach : 
$log_{3^2} \frac{1}{2} > 0$
$\Rightarrow \frac{1}{2} log_3 \frac{1}{2} >0$
$\Rightarrow \frac{1}{2} [ log_3 1 -log_3 2] >0                  $
$\Rightarrow \frac{1}{2} [ 0 - log_3 2] >0                   $
$\Rightarrow -\frac{1}{2} [  log_3 2] > 0                   $ { which is false}
Please suggest... thanks...
 [/INST] To evaluate $log_{3^2}\frac{1}{2}$ you solve $(3^2)^x=\frac{1}{2} \rightarrow xlog(3^2)=log(\frac{1}{2}) \rightarrow x =\frac{log(\frac{1}{2})}{log(3^2)}$, but $log(\frac{1}{2}) = -ln(2)$. Hence,  $log_{3^2}\frac{1}{2}$ is not greater than 0.
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