The sets of real values of x for which
$l o g_{2 x + 3} x^{2} < l o g_{2 x + 3} (2 x + 3)$
includes

$(\frac{- 3}{2}, - 1)$

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(-1, 0)

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(0, 3)

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$(- \infty, 0)$

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Solution

The correct options are
A
$(\frac{- 3}{2}, - 1)$

B
(-1, 0)

C
(0, 3)

The inequality (1) is equivalent to the following system of inequalities
$\begin{matrix} 0 < 2 x + 3 < 1 X^{2} > 2 x + 3 \end{matrix}} (1)$
$o r \begin{matrix} 2 x + 3 > 1 0 < x^{2} < 2 x + 3 \end{matrix}} (2) S o l v i n g (1) f i r s t, w e s e e t h a t i t i s e q u i v a l e n t t o - 3 / 2 < x < - 1 a n d (x - 3) (x + 1) > 0 \Rightarrow - 3 / 2 < x < - 1 a n d (x < - 1 o r x > 3) \Rightarrow - 3 / 2 < x < - 1 N e x t s o l v i n g (2), w e s e e t h a t i t i s e q u i v a l e n t t o x > - 1 a n d (x + 1) (x - 3) < 0 \Rightarrow - 1 < x < 3 A l s o, x \neq 0 H e n c e, x ϵ (- 3 / 2, - 1) \cup (- 1, 0) \cup (0, 3)$

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