How do you solve #e^(4-2x) = 45#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer JustMe314 Aug 11, 2018 #e^{4-2x}=45# #ln(e^{4-2x})=ln 45# #4-2x=ln45# #-2x=ln45-4# #2x=4-ln45# #x=\frac{4-ln45}{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 1632 views around the world You can reuse this answer Creative Commons License