Help solving the following equation? #log_7(x-2) + log_7(x+4)=2log_7 4# Please also check for extraneous solutions. Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer P dilip_k Dec 10, 2016 #log_7(x-2) + log_7(x+4)=2log_7 4# #=>log_7[(x-2)(x+4)]=log_7 4^2# #=>[(x-2)(x+4)]= 4^2# #=>x^2+2x-8-16=0# #=>x^2+2x-24=0# #=>x^2+6x-4x-24=0# #=>x(x+6)-4(x+6)=0# #=>(x+6)(x-4)=0# So #x= -6 and x= 4# for #x=-6->##x-2 and x+4# become negative, so their logarithm is not possible. That means #x=-6# is extraneous solutions. Hence # x= 4# is only solution 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 5820 views around the world You can reuse this answer Creative Commons License