How do you solve #logx-log6=2log4#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer sjc Oct 25, 2016 # x=96# Explanation: #logx-log6=2log4# #=>logx=log4^2+log6# #=>logx=log16+log6=log(16xx6)# #=>logx=log96# #:. x=96# 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 7277 views around the world You can reuse this answer Creative Commons License