How do you solve #4ln(2x+3)=11#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Jul 14, 2016 #x=6.3213# Explanation: As #4ln(2x+3)=11#, #ln(2x+3)=11/4=2.75# or #2x+3=e^(2.75)=15.6426# or #2x=15.6426-3=12.6426# or #x=12.6426/2=6.3213# 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 6305 views around the world You can reuse this answer Creative Commons License