How do you solve for x in #90e^(-.879x)=225#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A. S. Adikesavan May 19, 2016 #x=ln 0.4/0.879=-1.042424#, nearly. Explanation: Here, #e^0.879=90/225=0.4# Inversely, 0.879 x= ln 0.4 So, #x=ln 0.4/0.879=-1.042424#, nearly 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 1115 views around the world You can reuse this answer Creative Commons License