Equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution; then exhaustive set of values of $a$ is Equation: 

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

It has only one solution; then exhaustive set of values of $a$ is ??
I don't even know where to begin
The answer is :
$$(-13/4,-13/12) \cup [-1]$$
 A: Hint: $a^{\log_{a}(x)} = x$, so:
$$x^2+2ax = 4x-4a-13$$
$$\Rightarrow x^2+2(a-2)x+(4a+13) = 0$$
And use the quadratic formula.
A: Removing  logs assuming $y in log(y)\neq 0$  we get $x^2+2ax-4x+4a+13=0$ thus only one solution implying delta=0 thus $(2a-4)^2-16a-208=0$ thus $4a^2-16a+16-16a-208=0$ thus $4a^2-32a-192=0$ ie $a^2-8a-48=0$ 
A: $$\log(x^2+2ax)=\log(4x-4a-13)$$
 This implies that $$x^2+2ax=4x-4a-13$$
or $$x^2+2ax-4x+4a+13=0$$
or $$x^2+(2a-4)x+(4a+13)=0$$
Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = $0$.
Hence we get that $$(2a-4)^2=4\cdot 1 \cdot (4a+13)$$
or $$4a^2-16a+16=16a+52$$
or $$4a^2-32a-36=0$$
or $$a^2-8a-9=0$$
or $$(a-9)(a+1)=0$$
So the values of $a$ are $-1$ and $9$.
A: Since log is one-one function hence $$\log(x^2+2ax)=\log(4x-4a-13)$$
$$x^2+2ax=4x-4a-13$$
$$x^2+2(a-2)x+4a+13=0$$
The given equation will have one solution if & only if above quadratic equation has only one root i.e. determinant, $\Delta\equiv B^2-4AC=0$
$$(2(a-2))^2-4(1)(4a+13)=0$$
$$a^2-8a-9=0$$ $$a=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(-9)}}{2(1)}=\frac{8\pm 10}{2}$$$$\color{red}{a\in\{-1, 9\}}$$
A: The other solutions mostly agree that the quadratic has a double-zero if $a=-1,9$.
But if $a=9$, then $(x+7)^2=0$, and $x=-7$, so $\log(x^2+2ax)$ doesn't exist.
The other way to get one solution is when $4x-4a-13$ is positive for exactly one of the two possible values of $x$.
