What the Function?

$\begin{aligned} f(2) & = \log_a \pi \\ \implies a^{f(2)} & = \pi \\ e^{\ln a \cdot f(2)} & = \pi \\ \ln a \cdot f(2) & = \ln \pi \\ \ln a & = \ln \pi \cdot \frac {(\ln 2)^2 + \pi^2}{\ln \pi (\ln 2 - \pi i)} \\ & = \frac {(\ln 2)^2 + \pi^2}{\ln 2 - \pi i} \\ & = \frac {\left((\ln 2)^2 + \pi^2\right)(\ln 2 + \pi i)}{(\ln 2 - \pi i)(\ln 2 + \pi i)} \\ & = \frac {\left((\ln 2)^2 + \pi^2\right)(\ln 2 + \pi i)}{(\ln 2)^2 + \pi^2} \\ & = \ln 2 + \pi i \\ \implies a & = e^{\ln 2 + \pi i} = e^{\ln 2} e^{\pi i} = 2(-1) = \boxed{-2} \end{aligned}$

We have that $f(x) = \dfrac{\ln{\pi}}{(\ln{x})^2+\pi^2}\cdot (\ln{x}-\pi i) \quad \forall x \neq 0$ and $f(2) = \log_a\pi$ .

So $f(2) = \dfrac{\ln(\pi)}{(\ln{2})^2 + \pi^2} \cdot (\ln{2}- \pi i) = \dfrac{(\ln{2}-\pi i) \cdot \ln{\pi}}{(\ln{2}+\pi i)(\ln{2}-\pi i)}.$

Since ${(\ln{2})^2+\pi^2 = (\ln{2}+\pi i)(\ln{2}-\pi i)}\\ \implies f(2) = \dfrac{\ln{\pi}}{\ln{2}+\pi i} = \dfrac{\ln{\pi}}{\ln{(-2)}} = \log_{(-2)}\pi = \dfrac{\ln{\pi}}{\ln{(-2)}}$

$\therefore \dfrac{1}{\ln{(-2)}} = \dfrac{1}{\ln{a}} \implies \ln{(-2)} = \ln{a} \\ \therefore a = \boxed{-2}$ .