Solve the inequality ${log}_{2 x + 3} x^{2} < {log}_{2 x + 3} (2 x + 3)$ .

x \in (\frac{5}{2}, 1) \cup (- 1, 1)

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x \in (\frac{7}{2}, 1) \cup (- 1, 3)

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x \in (\frac{3}{2}, 1) \cup (- 1, 2)

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x \in (\frac{3}{2}, 1) \cup (- 1, 3)

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Solution

The correct option is C $x \in (\frac{3}{2}, 1) \cup (- 1, 3)$
This inequation is equivalent to the collection of the system
$⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} 2 x + 3 > 1 x^{2} < 2 x + 3 0 < 2 x + 3 < 1 x^{2} > 2 x + 3 \end{matrix}$
$\Rightarrow ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x > - 1 (x - 3) (x + 1) < 0 - \frac{3}{2} < x < - 1 (x - 3) (x + 1) > 0 \end{matrix}$
$\Rightarrow ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x > - 1 - 1 < x < 3 \Rightarrow - 1 < x < 3 - \frac{3}{2} < x < - 1 \Rightarrow - \frac{3}{2} < x < - 1 x < - 1 a n d x > 3 \end{matrix}$
Hence solution of the original ineqatuion is $x \in (\frac{3}{2}, 1) \cup (- 1, 3)$

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