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 $x=9$

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