How do you solve #log_2 (3-x) + log_2 (2-x) = log_2 (1-x)#?

Question

1720 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-22T07:30:08

#color(chocolate)(x = 2 +- i)#

#log_2 (3-x) + log_2(2-x) = log_2 (1-x)#

#log_2 ((3-x)(2-x)) = log_2 (1-x)#

#((3-x)(2-x)) = (1-x)#

#x^2 - 5x + 6 = 1 - x#

#x^2 - 4x + 5 = 0#

#x = (4 +- sqrt(16 - 20)) / 2#

#x = (4 +- sqrt-4) / 2#

#color(chocolate)(x = 2 +- i)#