# If $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution, find the exhaustive set of values of $a$

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

This question was asked at Equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution; then exhaustive set of values of $a$ is, but the answers there don't match my book's given answer $$\left(-{13}4,-\frac{13}{12}\right) \cup [-1].$$

My Approach:

For $$\log(4x-4a-13)$$ to be valid, $$4x-4a-13>0,$$ i.e., $$x>\frac{4a+13}{4}.$$

For $$\log(x^2+2ax)$$ to be valid, $$x^2+2ax>0,$$ i.e., $$x(x+2a)>0;$$ here we have two cases

• case 1 $$a>0$$

$$\implies$$ $$x\in(-\infty,-2a)\cup (0,\infty)$$

• case 2 $$a<0$$

$$\implies$$ $$x\in(-\infty,0)\cup (2a,\infty).$$

For one solution, $$x^2+2ax=4x-4a-13$$ and its discriminant equals $$0.$$ So, I got $$a=-1$$ and $$a=9.$$ 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 the other part of the given solution?

• The given answer is not correct since $a=-\frac{13}{4}$ and $a=-\frac{13}{12}$ have to be included. Commented Jan 11, 2022 at 14:53

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

• @ryang : Note that $p,q$ are the solutions of $(1)$. So, both $p^2+2ap=4p-4a-13$ and $q^2+2aq=4q-4a-13$ hold. Commented Jan 11, 2022 at 14:35
• @ryang : I know. Since we have $p^2+2ap=4p-4a-13$, $4p-4a-13\gt 0$ is equivalent to $p^2+2ap\gt 0$. Also, since $q^2+2aq=4q-4a-13$, $4q-4a-13\gt 0$ is equivalent to $q^2+2aq\gt 0$. Commented Jan 11, 2022 at 14:42
• Oh yes, of course, hehe. +1. Commented Jan 11, 2022 at 14:54
• A very well written answer! Commented May 21, 2023 at 9:23

Find the set of values of $$a$$ for which $$\log(x^2+2ax)=\log(4x-4a-13)$$ has only one solution.

\begin{align}&\log(x^2+2ax)=\log(4x-4a-13) \\\iff{}&x^2+(2a-4)x+(4a+13)=0\quad\text{and}\quad 4x-4a-13>0 \\\iff{}&x=2-a\pm\sqrt{(a+1)(a-9)}\quad\text{and}\quad x>a+\frac{13}4.\end{align}

Therefore, the given equation having exactly one solution is equivalent to the disjunction of these cases:

1. $$a=-1\quad$$ and $$\quad2-a\pm0>a+\dfrac{13}4:$$

$$\iff a=-1\quad\text{and}\quad a<-\frac58\\\iff a=-1$$

2. $$a=9\quad$$ and $$\quad2-a\pm0>a+\dfrac{13}4:$$

$$\iff a=9\quad\text{and}\quad a<-\frac58\\\iff \bot$$

3. $$2-a-\sqrt{(a+1)(a-9)}\le a+\dfrac{13}4\quad$$ and $$\quad2-a+\sqrt{(a+1)(a-9)}>a+\dfrac{13}4:$$

$$\iff$$ $$-\frac{13}4\le a\le -\frac{13}{12}.$$

So, this disjunction is equivalent to $$a=-1\quad\text{or}\quad-\frac{13}4\le a\le -\frac{13}{12}.$$ Hence, the required set is $$\left[-\frac{13}4,-\frac{13}{12}\right]\cup\{-1\}.$$ (As pointed out by mathlove, your book's given solution is not quite correct.)

An analogous exercise using the same solution structure:

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

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 $$\sqrt3\leq\sqrt{p^2-1}$$

(positive discriminant but only one positive solution),

that is, iff $$p\in\{1\}\cup[2,\infty).$$

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 $$4x-4a-13 \le 0$$

• I still don't understand why you did $x^2+2ax\leq 0$ and $4x-4a-13\leq 0$ Commented Jan 11, 2022 at 4:00
• @mathophile I did that because $\log(x)$ is not defined for $x \le 0$. Either $\log(x^2+2ax)$ is not defined (i.e. $x^2+2ax \le 0$) or $\log(4x-4a-13)$ is not defined (i.e. $4x-4a-13 \le 0$) Commented Jan 11, 2022 at 4:17
• @AnkitSaha Your argument is flawed: remember, if $\log(x^2+2ax)$ is undefined, then so is $\log(4x-4a-13),$ since they are set as equal. Commented May 21, 2023 at 8:28
• @ryang I perhaps didn't word my argument very clearly. What I meant is that even if the quadratic equation $x^2+2ax=4x-4a-13$ has two two distinct roots, say $\alpha$ and $\beta$ and if $\alpha^2 + 2a\alpha > 0$ but $\beta^2 + 2a\beta < 0$ then the equation $\log(x^2+2ax) = \log(4x-4a-13)$ will still have exactly one solution, namely $\alpha$ because it is not defined at $\beta$. So both roots of the quadratic equation do not have to be equal for the logarithmic equation to have a unique solution. Commented May 23, 2023 at 17:08