How do you solve #Log(3x+2)-log(2x-1)=log 2#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alan P. Nov 25, 2015 #x=4# Explanation: In general #log(a/b) = log(a) - log(b)# Therefore #color(white)("XXX")log(3x+2)- log(2x-1) = log((3x+2)/(2x-2))# So #color(white)("XXX")log(3x+2)-log(2x+1) = log(2)# #rArrcolor(white)("XXX")log((3x+2)/(2x-1))= log(2)# #rArrcolor(white)("XXX")(3x+2)/(2x-1)=2# #rArrcolor(white)("XXX")3x+2 = 4x-2# #rArrcolor(white)("XXX")x=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 5600 views around the world You can reuse this answer Creative Commons License