How do you solve #ln y = 2x + 4#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alvin L. Mar 4, 2018 #y=e^(2x+4)# Explanation: Assuming the goal is to isolate #y#, we can do so by raising both sides as a power of #e#, cancelling the natural logarithm: #e^ln(y)=e^(2x+4)# #y=e^(2x+4)# 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 4487 views around the world You can reuse this answer Creative Commons License