How do you solve #log (6x+2) - log (3x-6) = log 6 - log 2#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Sihan Tawsik Jan 28, 2016 20/3 Explanation: #log(6x+2)-log(3x-6)=log6-log2# #or,log((6x+2)/(3x-6))=log(6/2)# #or,(6x+2)/(3x-6)=3# [by, adding antilog to each side] #or,6x+2=9x-18# #or,3x=20# #or,x=20/3# 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 2596 views around the world You can reuse this answer Creative Commons License