Number of solutions of the equation ${log}_{x^{2} + 6 x + 6} ({log}_{2 x^{2} + 2 x + 3} (x^{2} - 2 x)) = 0$ is

4

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2

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1

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0

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Solution

The correct option is B $1$
${log}_{x^{2} + 6 x + 6} ({log}_{2 x^{2} + 2 x + 3} (x^{2} - 2 x)) = 0$

To find number of solutions
$∴ x^{2} + 6 x + 6 > 0 \Rightarrow x^{2} + 6 x + 6 + 3 > 0 + 3$
$\Rightarrow (x + 3)^{2} > 3$ ....(i)

Now, ${log}_{2 x^{2} + 2 x + 3} (x^{2} - 2 x) = 1$
$\Rightarrow x^{2} - 2 x = 2 x^{2} + 2 x + 3$
$\Rightarrow x^{2} + 4 x + 3 = 0 \Rightarrow (x + 1) (x + 3) = 0$
$\Rightarrow x = - 1, - 3$

To check now, for condition (i)

$(- 1 + 3)^{2} > 3 \Rightarrow 4 > 3 \to T r u e$
$(- 3 + 3)^{2} > 3 \Rightarrow 0 > 3 \to F a l s e$
$\Rightarrow x \neq - 3$
Also, $x^{2} - 2 x > 0 \Rightarrow x (x - 2) > 0$
$\Rightarrow x ϵ (- \infty, 0) \cup (2, \infty)$
$∴ x = - 1$ is a solution of the equation given.

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