Solve the following equation:
${log}_{(2 x + 3)} ({6 x}^{2} + 23 x + 21) = 4 - {log}_{(3 x + 7)} ({4 x}^{2} + 12 x + 9)$ . Find $- 4 x .$

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Solution

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

$⟹ {log}_{(2 x + 3)} [(2 x + 3) (3 x + 7)] + {log}_{(3 x + 7)} {(2 x + 3)}^{2} = 4$

$⟹ {log}_{(2 x + 3)} (2 x + 3) + {log}_{(2 x + 3)} (3 x + 7) + 2 {log}_{(3 x + 7)} (2 x + 3) = 4$

$⟹ 1 + {log}_{(2 x + 3)} (3 x + 7) + \frac{2}{{log}_{(2 x + 3)} (3 x + 7)} = 4$

Let ${log}_{(2 x + 3)} (3 x + 7) = t$

$∴ t + \frac{2}{t} = 3$

$\Rightarrow t^{2} - 3 t + 2 = 0$

$∴ t = 1, 2$
When $t = 1$

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

$\Rightarrow 3 x + 7 = 2 x + 3$

$\Rightarrow∴ x = - 4$

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

i.e. $x > - \frac{3}{2}$ and $x > - \frac{7}{3}$

Hence no solution for $t = 1$
When $t = 2$

$\Rightarrow {log}_{(2 x + 3)} (3 x + 7) = 2$

$\Rightarrow 3 x + 7 = {(2 x + 3)}^{2}$

$\Rightarrow {4 x}^{2} + 9 x + 2 = 0$

$∴ x = - 2$ and $x = - \frac{1}{4}$

But $x = - 2$ can not be the solution because $x > - \frac{3}{2}$ and $x > - \frac{7}{3}$

$∴ x = - \frac{1}{4}$

So, the given equation has only one solution $x = - \frac{1}{4}$

Thus, $- 4 x = - 4 (- \frac{1}{4}) = 1$ .

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