How do you solve #ln(x+1) - lnx = 1#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alan P. Dec 20, 2015 #x=1/(e-1)# Explanation: Given: #ln(x+1)-ln(x)=1# #ln((x+1)/x)=1# #e^(ln((x+1)/x))=e^1# #(x+1)/x=e# #x+1 = x*e# #x-x*e = -1# #x*(1-e)=-1# #x=1/(e-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 20404 views around the world You can reuse this answer Creative Commons License