# How do you solve for x in log_3(x^(2)-2x)=1?

Dec 31, 2015

${x}_{1} = - 1$ and ${x}_{2} = 3$

#### Explanation:

${\log}_{3} \left({x}^{2} - 2 x\right) = 1$
${x}^{2} - 2 x = {3}^{1}$
${x}^{2} - 2 x - 3 = 0$
$\Delta = 4 + 12 = 16$
$x = \frac{- b \pm \sqrt{\Delta}}{2} = \frac{2 \pm 4}{2}$ => ${x}_{1} = - 1$ and ${x}_{2} = 3$