<s>[INST] Real roots of the equation $\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$ 
Find the set of real roots of the equation$$\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$$

My Attempt
$$
2x+3>0, 5x+4>0, 2x+3,5x+4\neq1\implies x>-4/5\;\&\;x\neq -1\;\&\;x\neq -3/5
$$
$$
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)(2x+3)=1\\
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)-\log_{(2x+3)}(2x+3)=1\\
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)-1=1\\
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)=2\\
3\log_{(5x+4)}(2x+3)-\frac{1}{\log_{(5x+4)}(2x+3)}=2
$$
Set $y=\log_{(5x+4)}(2x+3)$
$$
3y^2-2y-1=0\implies y=\log_{(5x+4)}(2x+3)=1,\frac{-1}{3}
$$
Case 1:
$$
\log_{(5x+4)}(2x+3)=\frac{\log(5x+4)}{\log(2x+3)}=1\implies\log(5x+4)=\log(2x+3) \implies \boxed{x=\frac{-1}{3}}
$$
Case 2:
$$
\log_{(5x+4)}(2x+3)=\frac{\log(2x+3)}{\log(5x+4)}=\frac{-1}{3}\implies \color{red} ?
$$
My reference says $x=\frac{-1}{3}$ is the only solution. How do I prove it ?
Possible Solution
$$
y=\log_{(5x+4)}(2x+3)=\frac{-1}{3}<0\:,\quad x>-4/5=-0.8
$$
Case 1: $5x+4>1\implies x>-3/5=-0.6$
$$
0<2x+3<1\implies -3/2<x<-1\implies-1.5<x<-1\\
\text{Not Possible !}
$$
Case 2: $0<5x+4<1\implies -4/5<x<-3/5\implies-0.8<x<-0.6$
$$
2x+3>1\implies x>-1\\
\implies \boxed{x\in(-0.8,-0.6)}
$$
It seems like there could be a solution for "case 2" in my attempt ?
 [/INST] Since $10x^2+23x+12=(2x+3)(5x+4)$, my suggestion is to set $t=\log_{(2x+3)}(5x+4)$, so that
$$
\log_{(5x+4)}(2x+3)=\frac{1}{t}
$$
and the equation becomes
$$
\frac{3}{t}-1-t=1
$$
so $t^2+2t-3=0$ and $t=1$ or $t=-3$.
The first case yields $5x+4=2x+3$, hence $x=-1/3$.
The second case yields $5x+4=(2x+3)^{-3}$. If we consider the two functions $f(x)=5x+4$ and $g(x)=(2x+3)^{-3}$, we can see that
$$
f(-2)=-6,\quad g(-2)=-1,\qquad
f(-1)=-1,\quad g(-1)=1,\qquad
f(0)=4,\quad g(0)=1/27
$$
Hence the equation $f(x)=g(x)$ has two roots in the interval $(-2,0)$. For the one in the interval $(-2,-1)$, we have $f(x)=g(x)<0$, so this is not a solution.
For the one in the interval $(-1,0)$, we have $f(x)=g(x)>0$ and not $1$. So the solution is good for the problem.

</s>