Solution of $l o g_{x^{2} + 6 x + 8} l o g_{x^{2} + 2 x + 3} (x^{2} - 2 x) = 0 i s$

a natural number

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a negative integer

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-1

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none of these

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Solution

The correct options are
B
a negative integer

C
-1

We must have $x^{2} + 6 x + 8 > 0, 2 x^{2} + 2 x + 3 > 0 a n d x^{2} + 6 x + 8 \neq 1, 2 x^{2} + 2 x + 3 \neq 1 ∴ (x + 4) (x + 2) > 0 a n d, (x + 1 / 2)^{2} + \frac{5}{4} > 0 \Rightarrow x ϵ (- \infty, - 4) \cup (- 2, \infty) a n d x \neq - 3 \pm \sqrt{2} (1) a n d x^{2} - 2 x > 0 i . e ., x ϵ (- \infty, 0) \cup (2, \infty)$
The given equation can be written as
$l o g_{2 x^{2} + 2 x + 3} (x^{2} - 2 x) = 1 \Rightarrow x^{2} - 2 x = 2 x^{2} + 2 x + 3 \Rightarrow x^{2} + 4 x + 3 = 0 \Rightarrow x = - 1, - 3 x = - 3 d o e s n o t s a t i s f y (1) s o x = - 1.$

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Solution of $l o g_{x^{2} + 6 x + 8} l o g_{x^{2} + 2 x + 3} (x^{2} - 2 x) = 0 i s$

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