Question

Values of $x$ satisfying the equation ${\log }_{\left(2x+3\right)}\left(6x^2+23x+21\right)=4-{\log }_{\left(3x+7\right)}\left(4x^2+12x+9\right)$

• A. $-\frac{1}{3}$
• B. $-\frac{1}{4}$
• C. $4-\frac{1}{5}$
• D. $-\frac{1}{6}$

Answer 1

The correct option is B $-\frac{1}{4}$

Given equation is

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

$\Rightarrow {\log }_{(2x+3)}(6x^2+14x+9x+21)=4-{\log }_{(3x+7)}({\left(2x\right)}^2+2\times 2x\times 3x+{\left(3\right)}^2)$

$\Rightarrow {\log }_{(2x+3)}(2x+3)(3x+7)=4-{\log }_{(3x+7)}{(2x+3)}^2$

$\Rightarrow {\log }_{(2x+3)}(2x+3)+{\log }_{(2x+3)}(3x+7)=4-{\log }_{(3x+7)}{(2x+3)}^2$ $\therefore$$({\log }_{a}xy={\log }_{a}x+{\log }_{a}y)$

$\Rightarrow {\log }_{(2x+3)}(2x+3)+{\log }_{(2x+3)}(3x+7)=4-2{\log }_{(3x+7)}(2x+3)$ $\therefore$$({\log }_a{x}^n=n{\log }_{a}x)$

$\Rightarrow 1+{\log }_{(2x+3)}(3x+7)=4-2{\log }_{(3x+7)}(2x+3)$ $\therefore$$({\log }_{a}a=1)$

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

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

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

Then

$y=3-\frac{2}{y}$

$y^2=3y-2$

$y^2-3y+2=0$

By factorize

$(y-2)(y-1)=0$

When $(y-2)=0$ then $y=2$ and when $(y-1)=0$ then $y=1$

Put the value of $y$

For $y=2$

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

$(3x+7)={(2x+3)}^2$ $\therefore$$({\log }_{a}x=b)$ then $x=a^b$

$(3x+7)=(4x^2+12x+9)$

$\Rightarrow 3x+7=4x^2+12x+9$

$\Rightarrow 4x^2+12x+9-3x-7=0$

$\Rightarrow 4x^2+9x-2=0$

By factorize and we get,

$(4x+1)(x+2)=0$

$x=\frac{-1}{4},-2$

for $y=1$

${\log }_{(2x+3)}(3x+7)=1$ $\therefore$$({\log }_{a}x=b)$ then $x=a^b$

$(3x+7)=(2x+3)$

$\Rightarrow 3x+7-2x-3=0$

$\Rightarrow x=-4$

Hence the solutions are $(x=-2,-4,\frac{-1}{4})$.