Solve the following equation for x.
${log}_{(2 x + 3)} ({6 x}^{2} + 23 x + 21) + {log}_{(3 x + 7)} ({4 x}^{2} + 12 x + 9) = 4$
Find the value of $- 8 x$

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Solution

The given equation is ${log}_{(2 x + 3)} ({6 x}^{2} + 23 x + 21) + {log}_{(3 x + 7)} ({4 x}^{2} + 12 x + 9) = 4$
On rewriting the equation using factorization rules we get
${log}_{(2 x + 3)} (2 x + 3) (3 x + 7) + {log}_{(3 x + 7)} {(2 x + 3)}^{2} = 4$
$\Rightarrow {log}_{(2 x + 3)} (2 x + 3) {log}_{(2 x + 3)} (3 x + 7) + 2 {log}_{(3 x + 7)} (2 x + 3) = 4$
$\Rightarrow 1 + {log}_{(2 x + 3)} (3 x + 7) + \frac{2}{{log}_{(2 x + 3)} (3 x + 7)} = 4$
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 $x = - \frac{1}{4}$
$∴ x = - \frac{1}{4}$ (x=-2 does not lie in domain)
So, the given equation has only one solution $x = - \frac{1}{4}$
So, $- 8 x = - 8 (- \frac{1}{4}) = 2$
Hence $2$ is the required answer.

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