How do you solve e^x = π^(1-x)?

$x = \log \frac{\pi}{\log e + \log \pi}$
Well it is ${e}^{x} = {\pi}^{1 - x} \implies \log {e}^{x} = \log {\pi}^{1 - x} \implies x \log e = \left(1 - x\right) \log \pi \implies x \left(\log e + \log \pi\right) = \log \pi \implies x = \log \frac{\pi}{\log e + \log \pi}$