Help solving the following equation? #log_7(x-2) + log_7(x+4)=2log_7 4# Please also check for extraneous solutions.

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Answer 1 · 2016-12-10T13:29:46

#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