How do you solve #e^x = 22.8#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Michael Oct 19, 2015 You can solve it like this: Explanation: #e^x=22.8# Take natural logs of both sides: #lne^x=ln22.8# #xcancel(lne)=3.127# #x=3.127# 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 1496 views around the world You can reuse this answer Creative Commons License