How do you solve $e^x = π^(1-x)$ ?

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Answer 1 · 2015-09-09T21:27:28

$x=logpi/(loge+logpi)$

Well it is $e^x=pi^(1-x)=>loge^x=logpi^(1-x)=>xloge=(1-x)logpi=>x(loge+logpi)=logpi=>x=logpi/(loge+logpi)$

How do you solve e^x = π^(1-x)e^x = π^(1-x)?