a) $f(n) = O(n^{\log a}-e)$ for some $e>0$
b) $f(n) = \Theta(n^{\log a})$
c) $f(n) = \Omega(n^{\log a}+e)$ for some $e>0$
I think the last is true since limit of $f(n)/g(n)$ is infinity. So $f(n)$ is little omega of $g(n)$. So an $e$ must exist for tight bound. Am I correct? or I am missing something?
First two are incorrect since $lim (f(n)/n^{\log a})$ is infinity, it cannot be $\Theta(n^{\log a})$. Since its $\Omega$, not $\Theta$, it cannot be O.
*All $\log a$ are to the base $b; b>1$.
Yes since
$$\frac{n^{\log a}\cdot k \log n}{n^{\log a}}\to \pm \infty$$
the third option is the correct one.