How do you solve for x in #log_6x=2-log_6 4#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Somebody N. Jul 3, 2018 #color(blue)(x=9)# Explanation: #log_6(x)=2-log_6(4)# #log_6(x)+log_6(4)=2# #log(a)+log(b)=log(ab)# #log_6(4x)=2# Raising the base to these: #6^(log_6(4x))=6^2# #4x=6^2# #x=(6^2)/4=36/4=9# 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 2324 views around the world You can reuse this answer Creative Commons License