How do you solve log_3(2-3x) = log_9(6x^2 - 9x + 2)log3(23x)=log9(6x29x+2)?

1 Answer
Jul 31, 2018

x in CC - RR

Explanation:

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