How do you solve e^(5-9 x)=990e5−9x=990? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A. S. Adikesavan Mar 31, 2016 x=(5-ln 990)/9x=5−ln9909. Explanation: If e^m=nem=n, inversely, m = ln n Here, e^(5-9x)=990e5−9x=990. m=5-9x, n=990m=5−9x,n=990. Inversely, 5-9x=ln 9905−9x=ln990. 9x = 5-ln 9909x=5−ln990. x=(5-ln 990)/9x=5−ln9909. 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)=819x−4=81? How do you solve logx+log(x+15)=2logx+log(x+15)=2? How do you solve the equation 2 log4(x + 7)-log4(16) = 22log4(x+7)−log4(16)=2? How do you solve 2 log x^4 = 162logx4=16? How do you solve 2+log_3(2x+5)-log_3x=42+log3(2x+5)−log3x=4? See all questions in Logarithmic Models Impact of this question 1636 views around the world You can reuse this answer Creative Commons License