How do you solve #ln(2x+1) = 2 - ln(x)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer RedRobin9688 Jun 27, 2018 #ln(2x+1)=2-ln(x)# #e^(ln(2x+1))=e^(2-ln(x))# #2x+1=e^2/x# #2x^2+x=e^2# #2x^2+x-e^2=0# #x=(-1+-sqrt((1+8(e^2))))/4# #2*(-1-sqrt((1+8(e^2))))/4+1<0# Therefore #ln(2*(-1-sqrt((1+8(e^2))))/4+1)# is not defined and not a valid input for x. This means #x=(-1+sqrt((1+8(e^2))))/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 10469 views around the world You can reuse this answer Creative Commons License