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
 A: 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
$$\color{red}{a\in\bigg[-\frac{13}{4},-\frac{13}{12}\bigg]\cup [-1]}$$
A: 

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

the given answer is $(-13/4,-13/12) \cup [-1].$
I got $a=-1.$ But how to arrive at the other part of the solution?

$$\log(x^2+2ax)=\log(4x-4a-13)\tag 1$$





case
domain of equation $(1)$




1
$$a<-\frac{13}4$$
$$\left(a+\frac{13}4,0\right)\cup\left(-2a,\infty\right)$$


2
$$-\frac{13}4\leq a\leq-\frac{13}{12}$$
$$\left(-2a,\infty\right)$$


3
$$a>-\frac{13}{12}$$
$$\left(a+\frac{13}4,\infty\right)$$




\begin{align}&\log(x^2+2ax)=\log(4x-4a-13)\tag1
\\\iff&x^2+(2a-4)x+(4a+13)=0\tag2
\\\iff&x=2-a\pm\sqrt{(a+1)(a-9)}\end{align}
\begin{align}\text {eqn $(2)$ has a single real root}&\iff a=-1 \;\text{ or }\; a=9
\\\text {eqn $(2)$ has two real roots}&\iff a<-1 \;\text{ or }\; a>9\end{align}





values of the parameter $a$ for whicheqn $(2)$ has real root(s) $p,q$
does $\,p\,$ and/or $\,q\,$ lie in eqn $(1)$'s domain?




1
$$a<-\frac{13}4\quad(p<q)$$
$$p\in \left(a+\frac{13}4,0\right)\text{ and }\;q\in\left(-2a,\infty\right)$$


2
$$-\frac{13}4\leq a\leq-\frac{13}{12}\quad(p<q)$$
$$p\not\in\left(-2a,\infty\right)\text{ and }\;q\in\left(-2a,\infty\right)$$


3a
$$-\frac{13}{12}< a<-1\quad(p<q)$$
$$p,q\in \left(a+\frac{13}4,\infty\right)$$


3b
$a=-1\quad(p=q)$
$$p\in \left(a+\frac{13}4,\infty\right)$$


3c
$a=9\quad(p=q)$
$$p\not\in \left(a+\frac{13}4,\infty\right)$$


3d
$$a>9\quad(p<q)$$
$$p,q\not\in \left(a+\frac{13}4,\infty\right)$$




Hence, picking out subcases 2 and 3b, $$\text {eqn $(1)$ has exactly one solution}\iff a\in\left[-\frac{13}4,-\frac{13}{12}\right]\cup\{-1\}.$$

Addendum
Here's an ostensibly different problem that—unless solved geometrically—requires the same solution structure:

A triangle $PQR$ is such that $PQ = 2, QR = p$ and $\measuredangle RPQ = 30^\circ.$For which values of p is $PR$ uniquely determined?

Let $PR=q.\:$ By the Cosine Rule, $$p^2=q^2+2^2-2(2)(q)\cos30^\circ\\
q^2-2\sqrt3q+(4-p^2)=0\\
q=\sqrt3\pm\sqrt{p^2-1}$$ $q$ is uniquely-determined iff either

*

*$p^2-1=0\;$ (zero discriminant)

or

*

*$p^2-1>0$ and $\sqrt{p^2-1}\geq\sqrt3$
(positive discriminant but only one positive solution),
i.e., $$p\in\{1\}\cup[2,\infty).$$
A: The reason you didn't get the other part of the solution set is that you assumed the discriminant of $x^2+2ax=4x-4a-13$ must be zero for a unique solution. This is certainly one of the cases where there is one solution, but not the only case. The quadratic can have two roots, if one of the two roots is invalid because of the domain of $\log(x)$, even in this case, the equation will have only one solution.
$$ a^2-8a-9 >0 $$
$$ \implies a \in (-\infty,-1) \cup(9, \infty)$$
Now, for these values of $a$, check for the cases where exactly one of $\log(x^2+2ax)$ and $\log(4x-4a-13)$ is undefined, i.e. $x^2+2ax \le 0$ or $4x-4a-13 \le 0$
