How do you solve #ln x - ln (x-1) = 2#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Somebody N. Jul 9, 2018 #color(blue)(x=(-e^2)/(1-e^2)~~1.156517643)# Explanation: #lna-lnb=ln(a/b)# #lnx-ln(x-1)=2# #ln(x/(x-1))=2# #e^(ln(x/(x-1)))=e^2# #x/(x-1)=e^2# #x-e^2x=-e^2# #x(1-e^2)=-e^2# #x=(-e^2)/(1-e^2)~~1.156517643# 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 1787 views around the world You can reuse this answer Creative Commons License