How do you solve #log_8x+log_8(x+6)=log_8(5x+12)#?

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Answer 1 · 2016-07-12T07:50:47

#:. x=3#

#log_8x + log_8(x+6)= log_8(5x+12)#

#=>log_8(x*(x+6))= log_8(5x+12)#

#=>x(x+6)= 5x+12#

#=>x^2+6x-5x-12=0#

#=>x^2+4x-3x-12=0#

#=>x(x+4)-3(x+4)=0#

#=>(x+4)(x-3)=0#

#:.x=-4 and x=3#

But x=-4 is not possible bacause
#log("-ve number")# is undefined.