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

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Answer 1 · 2018-06-22T07:30:08

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

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

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

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

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

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

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

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

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

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