How do you solve #2e^(0.5x)=45#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Jul 29, 2016 #x=6.227# Explanation: As #2e^(0.5x)=45#, we have #e^(0.5x)=45/2=22.5#. Hence #0.5x=ln22.5# and #x=ln22.5/0.5=ln22.5/(1/2)=2xxln22.5=2xx3.1135=6.227# 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 5497 views around the world You can reuse this answer Creative Commons License