Question

How do you solve `e^x = π^(1-x)`?

Answer 1

`x=logpi/(loge+logpi)`

Explanation:

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)`