How do you solve #log_3(x^2-6x)=3#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Sep 7, 2016 #x=-3# or #x=9# Explanation: As #log_3(x^2-6x)=3#, we have #(x^2-6x)=3^3=27# or #x^2-6x-27=0# or #x^2-9x+3x-27=0# or #x(x-9)+3(x-9)=0# or #(x+3)(x-9)=0# and hence #x=-3# or #x=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 7661 views around the world You can reuse this answer Creative Commons License