How do you solve #log_3(x^2-6x)=3#?

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Answer 1 · 2016-09-07T15:23:54

#x=-3# or #x=9#

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#