TITLE: if equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution, then exhaustive set of value of $a$ is?
QUESTION [6 upvotes]: 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

REPLY [2 votes]: 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., $$p\in\{1\}\cup[2,\infty).$$