How do you solve log_3(2-3x) = log_9(6x^2 - 9x + 2)?

Jul 31, 2018

$x \in \mathbb{C} - \mathbb{R}$

Explanation:

${\log}_{3} \left(2 - 3 x\right) = \frac{{\log}_{3} \left(6 {x}^{2} - 9 x + 2\right)}{{\log}_{3} 9}$

$2 {\log}_{3} \left(2 - 3 x\right) = {\log}_{3} \left(6 {x}^{2} - 9 x + 2\right)$

${\log}_{3} {\left(2 - 3 x\right)}^{2} = {\log}_{3} \left(6 {x}^{2} - 9 x + 2\right)$

$4 - 12 x + 9 {x}^{2} = 6 {x}^{2} - 9 x + 2$

$2 - 3 x + 3 {x}^{2} = 0$

$\Delta = 9 - 4 \cdot 2 \cdot 3 = - 15 < 0$