From Wikipedia $\ln (\pi) $ is unknown to be transcendental.
$e^{(ie^{(\ln(\pi)})}=-1$
$i(e^{(\ln(\pi)})=i\pi$ is transcendental.
Due to the
Lindemann–Weierstrass theorem
any transcendental number raised to a power $y$ which yields a non transcendental number implies that $y$ is transcendental or zero
This implies $\ln (\pi)$ is transcedental if I am not mistaken? If not where is my error?
Best Answer
No, you have misunderstood Lindemann-Weierstrass. Lindemann-Weierstrass (actually Lindemann) says $e^a$ is transcendental when $a$ is nonzero and algebraic. This is certainly not the case when $e$ is replaced by an arbitrary transcendental number. For example, $2^{\sqrt{2}}$ is transcendental and $\sqrt{2}$ is algebraic, but $(2^{\sqrt{2}})^\sqrt{2} = 2$ is algebraic.