I should end up with a range for $a$, but I end up with a single value for $a$ after evaluating $Big$ $O$ and $Big$ $\Omega$.
Problem: Prove $\log_a(n)$ is $Big$ $\Theta$ $(\log_2(n))$. For which range of values of a is this true?
UPDATE: My work so far...
For $Big$ $O$:
$\log_a(n) \leq c * \log_2(n)$
$\frac{\log_2(n)}{\log_2(a)} \leq c * \log_2(n)$
$2^\frac{1}{c} \leq a$
For $Big$ $\Omega$:
$2^\frac{1}{c} \geq a$