How do you solve for x in #9e^x=107#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Sep 6, 2016 #x=2.4756# Explanation: As #9e^x=107# #e^x=107/9# Now taking natural log on both sides #x=ln107-ln9# = #4.6728-2.1972# = #2.4756# 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 1739 views around the world You can reuse this answer Creative Commons License