How do you solve #Ln (x) - Ln (x-1) = Ln (3)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer SagarStudy Dec 30, 2015 #x=3/2# Explanation: #ln(x)-ln(x-1)=ln3# We know that: #ln(y/z)=lny-lnz# Here #y=x and z=x-1# #implies ln(x/(x-1))=ln3# #implies (x/(x-1))=3# #implies x=3(x-1)# #implies x=3x-3# #implies 2x=3# #implies x=3/2# 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 3597 views around the world You can reuse this answer Creative Commons License