How do you solve ${log}_{3} (2 - 3 x) = {log}_{9} (6 x^{2} - 9 x + 2)$ $log_3(2-3x) = log_9(6x^2 - 9x + 2)$ ?

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Answer 1 · 2018-07-31T16:08:36

$x in CC - RR$

$log_3 (2 - 3x) = frac{log_3 (6x^2 - 9x + 2)}{log_3 9}$

$2 log_3 (2 - 3x) = log_3 (6x^2 - 9x + 2)$

$log_3 (2 - 3x)^2 = log_3 (6x^2 - 9x + 2)$

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

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

$Delta = 9 - 4 *2*3 = -15 < 0$

How do you solve log_3(2-3x) = log_9(6x^2 - 9x + 2)log3(2−3x)=log9(6x2−9x+2)log_3(2-3x) = log_9(6x^2 - 9x + 2)?