Value of x satisfying the equation
${log}_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - l o g_{(3 x + 7)} (4 x^{2} + 12 x + 9)$

- \frac{1}{3}

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- \frac{1}{4}

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- \frac{1}{5}

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- \frac{1}{6}

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Solution

The correct option is B $- \frac{1}{4}$
${log}_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - {log}_{(3 x + 7)} (4 x^{2} + 12 x + 9)$
$\Rightarrow {log}_{(2 x + 3)} (2 x + 3) (3 x + 7) = 4 - {log}_{(3 x + 7)} {(2 x + 3)}^{2}$
$\Rightarrow {log}_{(2 x + 3)} (2 x + 3) + {log}_{(2 x + 3)} (3 x + 7) = 4 - {log}_{(3 x + 7)} {(2 x + 3)}^{2}$
$\Rightarrow 1 + {log}_{(2 x + 3)} (3 x + 7) = 4 - {log}_{(3 x + 7)} {(2 x + 3)}^{2}$
Let $α = {log}_{(2 x + 3)} (3 x + 7) \Rightarrow \frac{1}{α} = {log}_{(3 x + 7)} {(2 x + 3)}^{2}$
$\Rightarrow 1 + α = 4 - \frac{2}{α}$
$\Rightarrow α + \frac{2}{α} = 3$
$\Rightarrow α^{2} - 3 α + 2 = 0$
$\Rightarrow (α - 1) (α - 2) = 0$
$\Rightarrow α = 1$ or $α = 2$
$\Rightarrow {log}_{(2 x + 3)} (3 x + 7) = 1$ or ${log}_{(2 x + 3)} (3 x + 7) = 2$
$\Rightarrow 3 x + 7 = 2 x + 3$ or $3 x + 7 = {(2 x + 3)}^{2}$
$\Rightarrow x = - 4$ or $3 x + 7 = {4 x}^{2} + 12 x + 9$
$\Rightarrow x = - 4$ or ${4 x}^{2} + 12 x + 9 - 3 x - 7 = 0$
$\Rightarrow x = - 4$ or ${4 x}^{2} + 9 x + 2 = 0$
$\Rightarrow x = - 4$ or $(4 x + 1) (x + 2) = 0$
$\Rightarrow x = - 4$ or $x = - \frac{1}{4}$ or $x = - 2$
When $x = - 4$ then $2 x + 3 = - 5 < 0$
Hence ${log}_{(2 x + 3)} (6 x^{2} + 23 x + 21)$ is not defined.
When $x = - \frac{1}{4}$ then $2 x + 3 = 2 \times - \frac{1}{4} + 3 = - \frac{1}{2} + 3 = \frac{5}{3} > 0$
Hence ${log}_{(2 x + 3)} (6 x^{2} + 23 x + 21)$ is defined.
When $x = - 2$ then $2 x + 3 = 2 \times - 2 + 3 = - 4 + 3 = - 1 < 0$
Hence ${log}_{(2 x + 3)} (6 x^{2} + 23 x + 21)$ is not defined.
$∴ x = - \frac{1}{4}$

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