$\sqrt{n} = \Omega(3^{\log{}n})$ is not true? Why?
We'll start with some arithmetic, and assume the base of the logarithm to be $x$. First, remember the log law:
$$
\log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)} 
$$
So
$$
\begin{array}{rl}
3^{\log_{x}(n)} &amp;= 3^{\frac{\log_{3}(n)}{\log_{3}(x)}}\\
&amp;=(3^{\log_{3}(n)})^{1/\log_{3}(x)}
&amp;=n^{1/\log_{3}(x)}
\end{array}
$$
Whether $n^{\frac{1}{2}}$ is in big-omega of this then depends on what $x$ is.

If we make the right and proper computer science assumption that the base 2 logarithm is the only logarithm (that was a joke of course), then $1/\log_{3}(x)=1/\log_{3}(2)\approx 1.6$, which is certainly bigger than $1/2$. 

You'd need $x\geq 9$ for the statement to be true.