Solve for $x$ the following equation:
$l o g_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - l o g_{(3 x + 7)} (4 x^{2} + 12 x + 9)$ .

- 4

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- 2

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- \frac{1}{4}

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n o n e

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Solution

The correct option is B $- \frac{1}{4}$
Given,
${log}_{(2 x + 3)} (6 x^{2} + 23 x + 21) = 4 - {log}_{(3 x + 7)} (4 x^{2} + 12 x + 9)$
or, ${log}_{(2 x + 3)} {(2 x + 3) (3 x + 7)} = 4 - {log}_{(3 x + 7)} (2 x + 3)^{2}$
or, $1 + {log}_{(2 x + 3)} {(3 x + 7)} = 4 - 2 {log}_{(3 x + 7)} (2 x + 3)$
or, ${log}_{(2 x + 3)} {(3 x + 7)} = 3 - 2 {log}_{(3 x + 7)} (2 x + 3)$
or, ${log}_{(2 x + 3)} {(3 x + 7)} + 2 {log}_{(3 x + 7)} (2 x + 3) = 3$ .......(1)
Let ${log}_{(2 x + 3)} (3 x + 7) = p$ .....(2) in (1) we get,
$p + \frac{2}{p} = 3$
or, $p^{2} - 3 p + 2 = 0$
or, $(p - 2) (p - 1) = 0$
or, $p = 2, 1$ .
If $p = 1$ from (2) we get,
${log}_{(2 x + 3)} (3 x + 7) = 1$
or, $2 x + 3 = 3 x + 7$
or, $x = - 4$ , this is not possible as base of logarithm cann't be negative.
Similarly for $p = 2$ from (2) we have,
${log}_{(2 x + 3)} (3 x + 7) = 1$
or, $(2 x + 3)^{2} = 3 x + 7$
or, $4 x^{2} + 12 x + 9 = 3 x + 7$
or, $4 x^{2} + 9 x + 2 = 0$
or, $(4 x + 1) (x + 2) = 0$
As $x$ cann't be $- 2$ so $x = - \frac{1}{4}$ .

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

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{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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