if equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution, then exhaustive set of value of $a$ is? 

if equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution, then exhaustive set of value of $a$ is?



Given answer is
$(-13/4,-13/12) \cup [-1]$

My Approach:
for $\log(x^2+2ax)$ to be valid $x^2+2ax>0$
Here we have to cases
case $1$ $a>0$
$x(x+2a)>0$
$\implies$ $x\in(-\infty,-2a)\cup (0,\infty)$
case $2$ $a<0$
$\implies$ $x\in(-\infty,0)\cup (2a,\infty)$
For $\log(4x-4a-13)$ to be valid $4x-4a-13>0$
$\implies$ $x>\frac{4a+13}{4}$
For one solution
$x^2+2ax=4x-4a-13$
and discriminant of above quadratic must be $0$
so i got two value of $a$ those are $a=-1, a=9$
but for $a=9$ I got $x=-7$ which not a valid solution because $\log(x^2+2ax)$ will be invalid.
But how to arrive at other part of solution?
Same question has been asked here but it doesn't solve my query Equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution; then exhaustive set of values of $a$ is

We want to find $a$ such that there is only one $x$ satisfying
$$x^2+2ax=4x-4a-13\tag1$$
$$x^2+2ax\gt 0\tag2$$
$$4x-4a-13\gt 0\tag3$$
The discriminant of $x^2+(2a-4)x+4a+13=0$ has to be non-negative, so it is necessary that $a\in (-\infty,-1]\cup [9,\infty)$.

*

*$a=-1$ is sufficient since $x=3$ is the only solution.


*$a=9$ is not sufficient since the equation has no solutions.


*For $a\in (-\infty,-1)\cup (9,\infty)$, let $p,q\ (p\lt q)$ be the solutions of $(1)$ where $$p=2-a-\sqrt{(a+1)(a-9)},\quad q=2-a+\sqrt{(a+1)(a-9)}$$ Then, one can see that
$$\begin{align}&\text{$p$ satisfies $(2)$ and $(3)$}
\\\\&\iff 4\bigg(2-a-\sqrt{(a+1)(a-9)}\bigg)-4a-13\gt 0
\\\\&\iff 4\sqrt{(a+1)(a-9)}\lt -8a-5
\\\\&\iff -8a-5\gt 0\quad \text{and}\quad  16(a+1)(a-9)\lt (-8a-5)^2
\\\\&\iff a\lt -\frac{13}{4}\quad \text{or}\quad -\frac{13}{12}\lt a\lt -1\tag4\end{align}$$
and that
$$\begin{align}&\text{$q$ satisfies $(2)$ and $(3)$}
\\\\&\iff 4\bigg(2-a+\sqrt{(a+1)(a-9)}\bigg)-4a-13\gt 0
\\\\&\iff 4\sqrt{(a+1)(a-9)}\gt 8a+5
\\\\&\iff 8a+5\leqslant 0\quad \text{or}\quad \bigg(8a+5\gt 0\quad \text{and}\quad 16(a+1)(a-9)\gt (8a+5)^2\bigg)
\\\\&\iff a\lt -1\tag5\end{align}$$What we want is $a$ such that only one of $(4)(5)$ is satisfied. So, $a\in [-\frac{13}{4},-\frac{13}{12}]$.
Therefore, the answer is
$$\color{red}{a\in\bigg[-\frac{13}{4},-\frac{13}{12}\bigg]\cup [-1]}$$
