How do you solve #4log x - 6 log (x+2)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Harish Chandra Rajpoot Jul 17, 2018 #\log({x^4}/{(x+2)^6})# Explanation: Given that #4\logx-6\log(x+2)# #=2(2\logx-3\log(x+2))# #=2(\log(x^2)-\log(x+2)^3)# #=2\log({x^2}/{(x+2)^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 4878 views around the world You can reuse this answer Creative Commons License