Consider the inequality
$ c_1\lg(n) \leq \lg(k) \leq c_2\lg(n),\text{ for } n \geq n_{0} $
With $c_1,c_2,n_0 > 0$, $\lg(k) = \Theta(\lg(n))$
By deriving the actual relationship of $k$ with $n$, we have
$ \Rightarrow 2^{c_1\lg(n)} \leq 2^{\lg(k)} \leq 2^{c_2\lg(n)}\\ \Rightarrow 2^{\lg(n^{c_1})} \leq 2^{\lg(k)} \leq 2^{\lg(n^{c_2})}\\ \Rightarrow n^{c_1} \leq k \leq n^{c_2} $
If $c_1 \neq c_2$, is it still possible to provide tight bounds (e.g. Theta notation) for $k$ given that $k = \Omega(n^{c_1})$ and $k = O(n^{c_2})$?