How do you solve #e^x = π^(1-x)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Konstantinos Michailidis · Stefan V. Sep 9, 2015 #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)# Answer link Related questions What is a logarithmic model? How do I use a logarithmic model to solve applications? What is the advantage of a logarithmic model? How does the Richter scale measure magnitude? What is the range of the Richter scale? How do you solve #9^(x-4)=81#? How do you solve #logx+log(x+15)=2#? How do you solve the equation #2 log4(x + 7)-log4(16) = 2#? How do you solve #2 log x^4 = 16#? How do you solve #2+log_3(2x+5)-log_3x=4#? See all questions in Logarithmic Models Impact of this question 4483 views around the world You can reuse this answer Creative Commons License