How do you solve # log_3(x^2) - log_3(x+3) =3#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer bp Dec 20, 2015 #x= (27+-9sqrt5)/2# Explanation: LHS= #log_3 ((x^2)/(x+3))#, Hence it is #log_3 ((x^2)/(x+3))#=3 #((x^2)/(x+3))= 3^3 =27# #x^2 -27x -81=0# #x= (27+-sqrt (729-324))/2# #x= (27+-9sqrt5)/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 1465 views around the world You can reuse this answer Creative Commons License