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

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Solution

$l o g_{3 x + 7} (9 + 12 x + 4 x^{2}) + l o g_{2 x + 3} (6 x^{2} + 23 x + 21) = 4$
$\Rightarrow l o g_{3 x + 7} (4 x^{2} + 6 x + 6 x + 9) + l o g_{2 x + 3} (6 x^{2} + 14 x + 9 x + 21) = 4$
$\Rightarrow l o g_{3 x + 7} [2 x (2 x + 3) + 3 (2 x + 3)] + l o g_{2 x + 3} [2 x (3 x + 7) + 3 (3 x + 7)] = 4$
$\Rightarrow l o g_{3 x + 7} (2 x + 3)^{2} + l o g_{2 x + 3} (2 x + 3) (3 x + 7) = 4$
$\Rightarrow 2 l o g_{3 x + 7} (2 x + 3) + l o g_{2 x + 3} (2 x + 3) + l o g_{2 x + 3} (3 x + 7) = 4$
$\Rightarrow 2 l o g_{3 x + 7} (2 x + 3) + l o g_{2 x + 3} (3 x + 7) = 3$
say $l o g_{3 x + 7} (2 x + 3) = t$
$\Rightarrow 2 t + \frac{1}{t} = 3$
$\Rightarrow 2 t^{2} - 3 t + 1 = 0$
$\Rightarrow 2 t^{2} - 2 t - t + 1 = 0$
$\Rightarrow 2 t (t - 1) - 1 (t - 1) = 0$
$t = 1 / 2$ or $t = 1$
$∴ l o g_{3 x + 7} (2 x + 3) = \frac{1}{2}$ or $l o g_{3 x + 7} (2 x + 3) = 1$
$2 x + 3 = \sqrt{3 x + 7}$ or $\Rightarrow 2 x + 3 = 3 x + 7$
$\Rightarrow 4 x^{2} + 9 + 12 x = 3 x + 7$ or $\Rightarrow x = - 4$
$\Rightarrow 4 x^{2} + 9 x + 2 = 0$
$\Rightarrow 4 x^{2} + 8 x + x + 2 = 0$
$\Rightarrow 4 x (x + 2) + 1 (x + 2) = 0$
$x = - 2$ or $r = - 1 / 4$
$∴ x = - 2, - 1 / 4 r - 4$
But only $x = - 1 / 4$ is acceptable.

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Similar questions

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

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l o g_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - l o g_{(3 x + 7)} (4 x^{2} + 12 x + 9)

l o g 3 x + 7 (9 + 12 x + 4 x^{2})

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

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

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