Question

If ${\log }_{2x+3}(6x^2+23x+21)=4-{\log }_{3x+7}(4x^2+12x+9)$, then value of $x$ is equal to

• A. $2$
• B. $-4$
• C. $\frac{-1}{4}$
• D. $1$

Answer 1

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

$6x^2+23x+21=(2x+3)(3x+7)$

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

Similarly, $4x^2+12x+9=(2x+3{)}^2$

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

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

$\Rightarrow t=4-\frac{2}{t}[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$

$t+\frac{2}{t}=3$

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

$t=2,1$

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

$x=-4$

But $2x+3>0\Rightarrow x>-\frac{3}{2}$ and $3x+7>0\Rightarrow x>-\frac{7}{3}$

Therefore x is not equal to -4.

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

$\Rightarrow x=-\frac{1}{4}$

$\therefore$ Only value of $x=-\frac{1}{4}$.