Solving ${log}_{2} (3 - x) + {log}_{2} (1 - x) < 3$ we get domain of x be $(m, n)$ . Find $m + n$ ?

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Solution

${log}_{2} (3 - x) + {log}_{2} (1 - x) < 3$
Clearly $x < 1$
$\Rightarrow {log}_{2} [(3 - x) (1 - x)] < 3$
$\Rightarrow (x - 1) (x - 3) < 2^{3} = 8$
$\Rightarrow x^{2} - 4 x - 5 = (x - 5) (x + 1) < 0$
Since, $x < 1$
Therefore, $x \in (- 1, 1)$
Hence, $m + n = 0$ .

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