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For which values of x does the inequality log base 1/2 of ((x² + 6x + 9) / (2(x + 1))) become less than −log base 2 of (x + 1)?

1. (-1, -1 + 2√2)
2. (1 − 2√2, 2)
3. (-1, ∞)
4. None of these

Correct answer: None of these

Solution

Since (x^2+6x+9)=(x+3)^2 and log_(1/2)(A)=-log_2(A), the inequality becomes log_2((x+3)^2/(2(x+1)))>log_2(x+1) with x>-1. This gives (x+3)^2>2(x+1)^2 -> x^2-2x-7<0 -> x in (1-2*sqrt(2), 1+2*sqrt(2)). Intersecting x>-1 yields (-1, 1+2*sqrt(2)) approx (-1, 3.83), which matches none of the options.

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