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

Sep 7, 2016

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

#### Explanation:

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

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

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

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

$x \left(x - 9\right) + 3 \left(x - 9\right) = 0$ or

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

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