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

1 Answer
Vinícius Ferraz
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#

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