How do you solve #e^(4x-1)=(e^2)^x #? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Konstantinos Michailidis · Stefan V. Sep 9, 2015 #x=1/2# Explanation: Well it is #e^(4x-1)=e^(2x)=>e^(4x-2x-1)=1=>e^(2x-1)=1=>2x-1=0=>x=1/2# 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 3469 views around the world You can reuse this answer Creative Commons License