Question

Solve the equation

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

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Answer 1

$\large\underline{\sf{Solution-}}$

Given Logarithmic equation is

$\rm \: log_{(2x + 3)}( {6x}^{2} + 23x + 21 ) = 4 - log_{3x + 7}( {4x}^{2} + 12x + 9 )$

Let first define the domain

2x + 3 > 0, it implies x > - 3/2

2x + 3 ≠ 1, it implies x ≠ - 1

3x + 7 > 0, it implies x > - 7/3

3x + 7 ≠ 1, it implies x ≠ - 2

Now, Consider

$\rm \: log_{(2x + 3)}( {6x}^{2} + 23x + 21 ) = 4 - log_{3x + 7}( {4x}^{2} + 12x + 9 )$

On splitting the middle terms, we get

$\rm \: log_{(2x + 3)}( {6x}^{2} + 9x + 14x + 21 ) = 4 - log_{3x + 7}( {4x}^{2} + 6x + 6x + 9 )$

$\rm \: log_{(2x + 3)}(3x(2x + 3) +7(2x + 3)) = 4 - log_{3x + 7}( 2x(2x + 3) + 3(2x + 3))$

$\rm \: log_{(2x + 3)}((2x + 3)(3x + 7)) = 4 - log_{3x + 7}( (2x + 3)(2x + 3))$

We know,

$\boxed{\sf{ \: \: log_{a}(ab) = 1 + log_{a}(b) \: \: }}$

So, using this result, we get

$\rm \:1 + log_{(2x + 3)}(3x + 7) = 4 - log_{3x + 7}(2x + 3)^{2}$

$\rm \:log_{(2x + 3)}(3x + 7) = 3 - 2log_{3x + 7}(2x + 3)$

We know,

$\boxed{\sf{ \: \: log_{x}(y) \: = \: \frac{1}{ log_{y}(x) } \: }}$

So, using this, the above can be rewritten as

$\rm \:log_{(2x + 3)}(3x + 7) = 3 - \dfrac{2}{log_{(2x + 3)}(3x + 7)}$

$\rm \:log_{(2x + 3)}(3x + 7) + \dfrac{2}{log_{(2x + 3)}(3x + 7)} - 3 = 0$

Let assume that

$\rm \: log_{(2x + 3)}(3x + 7) = y$

So, above expression can be rewritten as

$\rm \: y + \dfrac{2}{y} - 3 = 0$

$\rm \: \dfrac{ {y}^{2} + 2 - 3y}{y} = 0$

$\rm \: {y}^{2} - 3y + 2 = 0$

$\rm \: {y}^{2} - 2y - y + 2 = 0$

$\rm \: y(y - 2) - 1(y - 2) = 0$

$\rm \: (y - 2)(y - 1) = 0$

$\rm\implies \:y = 2 \: \: or \: \: y = 1$

So, Consider y = 2

$\rm \: log_{(2x + 3)}(3x + 7) = 2$

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

$\rm \: 3x + 7 = {4x}^{2} + 9 + 12x$

$\rm \: {4x}^{2} + 9 + 12x - 3x - 7=0$

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

$\rm \: {4x}^{2} +8x + x + 2=0$

$\rm \: 4x(x + 2) +1(x + 2) = 0$

$\rm \: (x + 2)(4x + 1) = 0$

$\rm\implies \:x = - 2 \: \: or \: \: x = - \dfrac{1}{4}$

Now, x = - 2 rejected because of domain.

Now, Consider y = 1

$\rm \: log_{(2x + 3)}(3x + 7) = 1$

$\rm \: 3x + 7 = 2x + 3$

$\rm\implies \:x = - 4$

Now, x = - 4 is rejected as x > - 7/3

So,

$\rm\implies \:\: x = - \dfrac{1}{4}$