Find the exhaustive set of values of 'a' such that log(x²+2ax)=log(8x-6a-3) has only one solution for x.

Question

Now I've tried to look up for the solution but haven't got any. One solution of a similar question makes the log functions on both sides disappear and simply makes the two brackets equal, x²+2ax=8x-6a-3 and goes on to solve the equation. The values of a obtained via this method are 1 and 13. Now it can't be 13 because if a is 13 then x would be (-9). This can't be because if a equals 13 and x equals -9 then (8x-6a-3) becomes -153. Log of -153 is undefined. a=1 is a solution but I don't know how to solve for other values. The answer to this is open interval (-1/2, -3/22) and {1}. A detailed solution is much appreciated.

Answer 1

The solutions $x$ of $x^2 + 2ax = 8x-6a-3$ are $$x=4-a \pm \sqrt{a^2-14a + 13}$$

There are no solutions if the discriminant is negative, so we must have $a^2-14a + 13 = (a-1)(a-13) \ge 0$, which is true if and only if $a \le 1$ or $a \ge 13$.

As you've discovered, there is only one solution for this equation when $a =1$ or $a = 13$. But in addition to this equation being true, the logarithms of each side must exist, which means that $8x > 6a + 3$ must be true for $x$ to be a solution. For $a = 1, x = 3$, which meets the condition. But when $a = 13,x = -9$, which does not. So of those two, only $a=1$ provides a solution to the logarithmic equation.

But when the quadratic equation has two solutions, it can be case that one of those solutions satisfies $8x > 6a+3$, while the other does not. Then the logarithmic equation will have only one solution, as required. If for some value of $x, 8x > 6a+3$ holds, then it will also hold for all greater values of $x$. Thus it must be the greater of the two solutions for which $8x > 6a+3$, while it is false for the lower solution. Thus we are looking for values of $a$ such that

$$8\left[4-a -\sqrt{a^2-14a + 13}\right] \le 6a+3\quad\text{ and }\quad 8\left[4-a +\sqrt{a^2-14a + 13}\right] > 6a+3\\ 8\sqrt{a^2-14a + 13} \ge 29-14a\quad\text{ and }\quad 8\sqrt{a^2-14a + 13} > 14a-29$$ As long as the square root is real, it is $\ge 0$, so either $29 - 14a > 0$ (whereby the RH inequality is automatically satisfied as $14a -29 < 0$) and $8\sqrt{a^2-14a + 13} = 29-14a$, or else $8\sqrt{a^2-14a + 13} > |14a-29|$.

$29 - 14a > 0$ holds when $a < 1$, but not when $a > 13$, so if $8\sqrt{a^2-14a + 13} = 29-14a$, we must have $a < 1$. Squaring both sides, $$64a^2 - 896a + 832 = 196a^2 - 1624a + 841\\132a^2 -728a + 9 = 0$$ of which, only the root $$a= \frac{728 - \sqrt{525232}}{264} = \frac{182 - \sqrt{32827}}{66} \approx 0.0124$$ is less than $1$. When $a$ has this value, the lower of the two solutions $x$ of the quadratic is $\frac{6a+3}8$, which leaves $\log{8x-6a-3}$ undefined, while the other solution is greater, having well-defined logarithm.

The other condition for one root is $8\sqrt{a^2-14a + 13} > |14a-29|$, which again by squaring both sides leads to the condition $132a^2 -728a + 9 < 0$. This is true for $a$ approximately between $0.0124$ and $5.5$. But only the portion of that interval $\le 1$ lies in the range where $a^2 -14a + 13 \ge 0$. Thus there is only one solution $x$ when $a$ in the interval $$\left[\frac{182 - \sqrt{32827}}{66}, 1\right]$$

Answer 2

I think the method for this question is right here but slight mistake in this line $29 -14a > 0$. It holds when $a<1$ but not when $a>13$, so if $\sqrt{8}a^2−14a+13 = 29−14a$, we must have $a<1$. Squaring both sides: $64a^2 − 896a + 832 = 196a^2 − 1624a + 841$.

Correct is 64a^2 − 896a + 832 = 196a^2 − 812a + 841$.

Thats all.

Answer 3

You basically need study the number of solutions of the system $$ \begin{cases} x^2+2ax = 8x-6a-3\\ 8x-6a-3>0. \end{cases} $$ Solving the first equation for $a$ and replacing it in the inequality yields the condition on $x$ $$ -3<x<0 \lor x > \frac3{11}.\tag{1}\label{1} $$ Now our aim becomes that of determining when the quadratic equation $$x^2+2x(a-4)+6a+3=0\tag{2}\label{2}$$ has a solution satisfying \eqref{1}.

In order for the quadratic \eqref{2} to have at least one solution, we need $a\leq 1 \lor a\geq 13$.

However, for $a\geq 13$ the solutions are both less then $-3$ (can you say why?), thus condition \eqref{1} is not satisfied.

Let us consider now the case $a\leq 1$. Note that the larger solution of the quadratic is decreasing with $a$. On the other way around, the smaller one is an increasing function of $a$.

So while $a$ increases form $-\infty$ to $1$, the larger solution goes from $+\infty$ to $3$ (why?), whereas the smaller one goes from $-3$ (again why?) to $3$.

By easy direct check the smaller solution is equal to $0$ when $a=-\frac12$. Thus, from what we noted above, for $a \leq -\frac12$ there are two solution to the original system, one between $-3$ and $0$, and the other one greater than $3$.

Again by direct check, the smaller solution is equal to $\frac3{11}$ when $a=-\frac3{22}$. Hence for $-\frac12 < a < -\frac3{22}$ the solutions of the system is one (precisely, the larger solution of the quadratic, because the smaller one falls into the interval $\left[0,\frac3{11}\right]$).

For $-\frac{3}{22}\leq a <1$ again we have two solution to the original system (both greater than $\frac3{11}$).

Finally, we have one solution again for $a=-1$.

So your original equation has exactly one solution when $$\boxed{-\frac12 < x < -\frac3{22} \lor a=1}.$$

Answer 4

You can do it by the following method:

1. Firstly make a quadratic equation: We have log(x²+2ax) = log(8x-6a-3) Remove log from both sides, on rearranging, x²+(2a-8)x+6a+3=0
2. Now apply discriminant=0 for only 1 root of quadratic equation. (2a-8)²-4(1)(6a+3)=0 On simplifying, a²-14a+13=0 a=13 or a=1 Putting a=13,our quadratic will become x²+18x+81=0 from here x=(-9) which makes 8x-6a-3, a negative quantity and hence log of it will be undefined.

Hence a=1 is a solution.

3. Think, if our quadratic i.e, x²+(2a-8)x+6a+3=0 will have only one root then a=1 but if our quadratic has 2 roots and only one root satisfies the condition of the quantity in log is positive, then x has only one solution and it does not violate the question's guidelines. Consider any of the quantity in log (I am taking x²-2ax , you can also take 8x-6a-3, and with the same method, answer will be same) Let the roots of this equation be p and q Sum of roots = -(2a-8) = 8-2a Product of roots = 6a+3

4. Now f(x) = x²-2ax f(p)×f(q)<0 [because any one of f(p) or f(q) is negative because any one of p or q is making negative and hence not satisfying the condition of log]

(p²-2ap)×(q²-2aq)<0 On expanding, use p+q as sum and pq as product of roots of our quadratic

On simplifying,you will get, (6a+3)(22a+3)<0 Here is the answer of this inequality, a belongs to (-1/2, -3/22) also we found that a = 1

So the answer is a belongs to (-1/2, -3/22)U{1}