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 ?
Best Answer
One can show why this case yields exactly one solution, as follows.
First, the expression on $\text{LHS}$ of the equation $$\log_{(5x+4)}{(2x+3)}=-\frac 13$$ uniquely defines a real number provided that $1\ne 5x+4>0$ and $2x+3>0.$ Thus, any solution of the equation must also satisfy the conditions $$x>-0.8, x\ne -0.6.$$
Having done that, I now show that this equation has exactly one root satisfying these conditions, thereby confirming the claim.
This equation is, by definition, equivalent to the polynomial equation $$40x^4+212x^3+414x^2+351x+107=p(x)=0.$$ First, note that this has no real roots for $x\ge 0$ since then $p(x)>0.$ Now since $p(-1)<0,$ it follows that there is at least one root in $(-1,0).$ We show that there is in fact a unique root in this interval. To see this, take the first derivative of $p(x),$ which gives $$p'(x)=160x^3+636x^2+828x+351.$$ The second derivative is then given by $$p''(x)=40x^2+106x+69,$$ whose discriminant is $14,$ whence its roots are $-1.5,-1.15,$ both outside of the interval $(-1,0).$ It follows that $p''>0$ in this interval. Therefore, the first derivative $p'(x)$ increases for all $x\in(-1,0).$ This means $p$ is convex in that interval; together with the negativity of $p$ at $-1,$ it implies that $p$ has at most one root in this interval. Finally, I show that this root must lie in $(-0.8,-0.6),$ finishing off the problem.
Now, we have that $p(-0.6)>0.$ Furthermore, we have that $p(-0.8)<0.$ Thus, we have confirmed the claim that the unique root in $(-1,0)$ falls within the subinterval $(-0.8,-0.6),$ which lies within the permitted range $-0.6\ne x>-0.8.$ Since there are no roots for $x\ge -0.6,$ it follows that no other root of $p(x)=0$ falls in the permitted range. This completes the proof.