How to solve the equation $\log_{2x+3}(6x^2+23x+21)=4-\log_{3x+7}(4x^2+12x+9)$? How to solve the equation $\log_{2x+3}(6x^2+23x+21)=4-\log_{3x+7}(4x^2+12x+9)$ ?
Can someone please tell me a few steps as to how to approach these category of problems? I know $2x+3>0$ and $3x+7>0$ is a must.

HINT:
Using Laws of Logarithms,
$$\log_{2x+3}(6x^2+23x+21)=\log_{2x+3}(2x+3)+\log_{2x+3}(3x+7)=1+\log_{2x+3}(3x+7)$$
Let $\log_{2x+3}(3x+7)=a$
$$\log_{3x+7}(4x^2+12x+9)=\log_{3x+7}(2x+3)^2=2\log_{3x+7}(2x+3)=\dfrac2a$$ as $\log_{3x+7}(2x+3)\cdot\log_{2x+3}(3x+7)=1$ 
