# Logarithm problem : Prove that l o g <mrow class="MJX-TeXAtom-ORD">

Logarithm problem : Prove that $lo{g}_{{3}^{2}}\frac{1}{2}>0$
My approach :
$lo{g}_{{3}^{2}}\frac{1}{2}>0$
$⇒\frac{1}{2}lo{g}_{3}\frac{1}{2}>0$
$⇒\frac{1}{2}\left[lo{g}_{3}1-lo{g}_{3}2\right]>0$
$⇒\frac{1}{2}\left[0-lo{g}_{3}2\right]>0$
$⇒-\frac{1}{2}\left[lo{g}_{3}2\right]>0$ { which is false}
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g2joey15
It is hard to prove something that isn't true. Alpha shows ${\mathrm{log}}_{9}\frac{1}{2}<-0.3$ so you aren't even close. Generally, logs of things less than 1 are negative.
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pokoljitef2
To evaluate $lo{g}_{{3}^{2}}\frac{1}{2}$ you solve $\left({3}^{2}{\right)}^{x}=\frac{1}{2}\to xlog\left({3}^{2}\right)=log\left(\frac{1}{2}\right)\to x=\frac{log\left(\frac{1}{2}\right)}{log\left({3}^{2}\right)}$, but $log\left(\frac{1}{2}\right)=-ln\left(2\right)$. Hence, $lo{g}_{{3}^{2}}\frac{1}{2}$ is not greater than 0.