Question

$log3x+7(9+12x+4x^2)$ +$log2x+3$ $(6x^2+23x+21)$ =4

Answer 1

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

Consider $4x^2+12x+9={\left(2x\right)}^2+2\times 2x\times 3+3^2={(2x+3)}^2$

And $6x^2+23x+21=0$

$\Rightarrow 6x^2+9x+14x+21=0$

$\Rightarrow 3x(2x+3)+7(2x+3)=0$

$\Rightarrow (3x+7)(2x+3)=0$

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

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

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

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

Let $\alpha ={\log }_{(3x+7)}(2x+3)$ then ${\log }_{(2x+3)}(3x+7)=\frac{1}{\alpha }$

$\Rightarrow 2\alpha +\frac{1}{\alpha }=3$

$\Rightarrow 2{\alpha }^2-3\alpha +1=0$ is quadratic in $\alpha$

$\Rightarrow 2{\alpha }^2-2\alpha -\alpha +1=0$

$\Rightarrow 2\alpha (\alpha -1)-1(\alpha -1)=0$

$\Rightarrow (\alpha -1)(2\alpha -1)=0$

$\Rightarrow \alpha =1,\alpha =\frac{1}{2}$

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

$\Rightarrow 2x+3=3x+7,2x+3={(3x+7)}^{\frac{1}{2}}$

$\Rightarrow 2x-3x=7-3,{(2x+3)}^2=3x+7$

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

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

$\Rightarrow x=-4,{4x}^2+8x+x+2=0$

$\Rightarrow x=-4,4x(x+2)+(x+2)=0$

$\Rightarrow x=-4.(4x+1)=0,(x+2)=0$

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

For $x=-2,-4$ the base will become negative.

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