How do you solve #ln(x+1) - lnx = 2#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Ahmemaru May 6, 2018 #x = 1/(e^2 - 1)# Explanation: #ln(x+1)-lnx = 2# #ln((x+1)/x) = ln(e^2)# #cancel(ln)((x+1)/x) = cancel(ln)(e^2)# #(x+1)/x = e^2# #x+1 = xe^2# #1 = xe^2 - x# common factor #1 = x(e^2 - 1)# #x = 1/(e^2 - 1)# 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 2379 views around the world You can reuse this answer Creative Commons License