How do you solve #2e^x-5e^x-3=0#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer vince Jul 23, 2016 #2e^x -5e^x# is #-3e^x#, therefore your equation is equivalent to #e^x =-1# : no solution, because #e^x>0# for any real #x#. 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 2019 views around the world You can reuse this answer Creative Commons License