How do you solve #log_2(3x-4) +log_2(5x-2)=4#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer ali ergin May 29, 2016 #x={-4/15,2}# Explanation: #log_a b+log_a c=log_a (b*c)# #log_2(3x-4)+log_2(5x-2)=4# #log_2(3x-4)(5x-2)=4# #log_a b=x" ; "b=a^x# #(3x-4)(5x-2)=2^4# #15x^2-6x-20x+8-16=0# #15x^2-26x-8=0# #(x-2)(15x+4)=0# #(x-2)=0" ; " rarr" "x=2# #(15x+4)=0" ; "rarr " "x=-4/15# #x={-4/15,2}# 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 2983 views around the world You can reuse this answer Creative Commons License