How do you solve #e^(5-9 x)=990#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A. S. Adikesavan Mar 31, 2016 #x=(5-ln 990)/9#. Explanation: If #e^m=n#, inversely, m = ln n Here, #e^(5-9x)=990#. #m=5-9x, n=990#. Inversely, #5-9x=ln 990#. #9x = 5-ln 990#. #x=(5-ln 990)/9#. 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 1379 views around the world You can reuse this answer Creative Commons License