How do you solve #e^(4x - 6) - 8 = 11212#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer José F. Feb 7, 2016 #x=ln(root(4)(11220e^6))# Explanation: #e^(4x-6)-8=11212# #e^(4x-6)=11220# #4x-6=ln(11220)# #4x=ln(11220)+6# #x=(ln(11220)+6)/4# 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 1403 views around the world You can reuse this answer Creative Commons License