$$ \log_{3x+7}{(9+12x+4x^2)}+ \log_{2x+3}{(6x^2+23x+21)} \ge 4$$ The logarithmic inequality is defined for: $ x \in (-3/2, -1) \cup(-1, \infty)$.
First, I supposed that my solutions are in the interval $(-3/2, -1)$. Following this interval for which the logarithmic function $t$ are decreasing and multiplying both sides of the inequality by $\log_{2x+3}{(3x+7)}$, I obtained: $$ 2+ [1+\log_{2x+3}{(3x+7)}-\log_{2x+3}{6}]\cdot\log_{2x+3}{(3x+7)} \le4\log_{2x+3}{(3x+7)}\\ \log_{2x+3}{(3x+7)}=t<0\\t^2-(3+\log_{2x+3}{6})t+2 \le0$$ And here I got stuck. This inequailty is intended to be solved witouth a calculator, is it possible to continuie doing the inequality withouth it? And what I did wrong?
Hint:
$$9+12x+4x^2=(2x+3)^2$$
$$ 6x^2+23x+21=(2x+3)(3x+7)$$
$$4\le\log_{3x+7}(2x+3)^2+\log_{2x+3}(2x+3)(3x+7)=2\log_{3x+7}(2x+3)+1+\log_{2x+3}(3x+7)$$
As $\log(y)$ is real for $y>0,$
we need $2x+3, 3x+7>0\implies x>$max$(-3/2,-7/3)=-3/2$
Now if $\log_{3x+7}(2x+3)=a,$
$$4\le2a+1+\dfrac1a$$
$$\iff0\le2a+\dfrac1a-3=\dfrac{2a^2-3a+1}a=\dfrac{(2a-1)(a-1)}a$$
If $\dfrac{(2a-1)(a-1)}a=0, a=1$ or $\dfrac12$
Else we need $a(2a-1)(a-1)>0$