How can I find $\Theta(log(m_1)+...+log(m_k))$ as related to $m$? given: 
$$m_1+m_2+...+m_k=m$$
How can I find $\Theta(log(m_1)+...+log(m_k))$ as related to $m$?  

I know that i can doing that: $O(log(m_1)+...log(m_k))=O(log(m)+...+log(m))=O(k \cdot log(m))$ , but can I find something without $k$ ?
The $\log$ function is concave so, by Jensen's inequality,
$$\frac{1}{k}\sum_{i=1}^k\log m_i \leq \log\Big(\frac1k\sum_{i=1}^k m_i\Big)$$
i.e.,
$$\sum_{i=1}^k \log m_i \leq k\log\frac{m}{k} = k\,(\log m - \log k)\,,$$
which is basically what you got by bounding $\log m_i\leq \log m$.  We can't do any better than this because we could have $m_1 = \dots = k_m = \tfrac{m}{k}$, in which case $\sum_i\log m_i = k\log\tfrac{m}{k}$.

However, if $k$ is a fixed constant, then $k\log\tfrac{m}{k}=O(\log m)$.