If ${log}_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - {log}_{(3 x + 7)} (4 x^{2} + 12 x + 9)$ , then find the value of $4 x + 1$ .

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Solution

Given, ${log}_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - {log}_{(3 x + 7)} (4 x^{2} + 12 x + 9)$

$\Rightarrow {log}_{(2 x + 3)} (2 x + 3) (3 x + 7) = 4 - {log}_{(3 x + 7)} {(2 x + 3)}^{2} \Rightarrow 1 + {log}_{(2 x + 3)} (3 x + 7) = 4 - 2 {log}_{(3 x + 7)} (2 x + 3)$

Put ${log}_{(2 x + 3)} (3 x + 7) = y$

$∴ y + \frac{2}{y} - 3 = 0$

$\Rightarrow y^{2} - 3 y + 2 = 0 \Rightarrow (y - 1) (y - 2) = 0$

$\Rightarrow y = 1$ or $y = 2$

$\Rightarrow {log}_{(2 x + 3)} (3 x + 7) = 1$ 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 $4 x^{2} + 9 x + 2 = 0$

$\Rightarrow x = - 4$ or $(4 x + 1) (x + 2) = 0$

$∴ - 2, - 4, - \frac{1}{4}$ ...(i)

But, log exists only when $6 x^{2} + 23 x + 21 > 0$

$4 x^{2} + 12 x + 9 > 0$

$2 x + 3 > 0$ and $3 x + 7 > 0$

$\Rightarrow x > - \frac{3}{2}$ ...(ii)

$∴ x = - \frac{1}{4}$ is the only solution.

$\Rightarrow 4 x + 1 = - 1 + 1 = 0$

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