So $a = \log_n{a}$? By definition of logs, $n^a = a$. So $n = \sqrt[a]{a}$. For example, we could have $\log_{\sqrt{2}}{2} = 2$, which makes sense. EDIT: okay, so ALL numbers. This would be satisfied by any $n$ in the form $\sqrt[a]{a}$.

The desired base of logarithm ranges from 0 to $e^{\frac{1}{e}}$ except 1.