How do you solve ${log}_{3} (x^{2} - 6 x) = 3$ $log_3(x^2-6x)=3$ ?

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Answer 1 · 2016-09-07T15:23:54

$x = - 3$ $x=-3$ or $x = 9$ $x=9$

As ${log}_{3} (x^{2} - 6 x) = 3$ $log_3(x^2-6x)=3$ , we have

$(x^{2} - 6 x) = 3^{3} = 27$ $(x^2-6x)=3^3=27$ or

$x^{2} - 6 x - 27 = 0$ $x^2-6x-27=0$ or

$x^{2} - 9 x + 3 x - 27 = 0$ $x^2-9x+3x-27=0$ or

$x (x - 9) + 3 (x - 9) = 0$ $x(x-9)+3(x-9)=0$ or

$(x + 3) (x - 9) = 0$ $(x+3)(x-9)=0$ and hence

$x = - 3$ $x=-3$ or $x = 9$ $x=9$

How do you solve log3(x2−6x)=3log_3(x^2-6x)=3?