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

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Answer 1 · 2018-06-22T07:30:08

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

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${log}_{2} (3 - x) + {log}_{2} (2 - x) = {log}_{2} (1 - x)$ $log_2 (3-x) + log_2(2-x) = log_2 (1-x)$

${log}_{2} ((3 - x) (2 - x)) = {log}_{2} (1 - x)$ $log_2 ((3-x)(2-x)) = log_2 (1-x)$

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

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

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

$x = \frac{4 \pm \sqrt{16 - 20}}{2}$ $x = (4 +- sqrt(16 - 20)) / 2$

$x = \frac{4 \pm \sqrt{-} 4}{2}$ $x = (4 +- sqrt-4) / 2$

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

How do you solve log2(3−x)+log2(2−x)=log2(1−x)log_2 (3-x) + log_2 (2-x) = log_2 (1-x)?