1. ## Logarithmic equation

Greetings,
The logarithmic equation below is given and it is required to find all values of the parameter a such that the equation has a single solution.

What puzzles me is that I believe first of all we would need to ensure that these logarithms exist so I look for all values of a for which the arguments of the logarithms are positive and get that a ranges from (0;1).

Now when I equal the discriminant to 0, I get a = 1, which should not be amonst the searched values of a, but is both because when I replace it in the equation I get truth and because one of the answers in the textbook is 1. Additionally, the other one states that every a < -1/3 works, which is in complete inconsistence with what I have for a.

2. First the argument of a logarithm function must be positive.
$a\in(0,1)$ insures that is true for both sides.

Second, the logarithm function is injective (one-to-one) on its domain.
Therefore solve $2x^2-4(a-2)+8+2a=x^2-2ax+a$.

3. Yes, according to this a = 1 should not be a solution, but it is.
Moreover, I think both arguments exist for a = 1.. ?
a = 1 or a < -1/3
Neither of which fit in (0;1).

4. Originally Posted by Logic
Yes, according to this a = 1 should not be a solution, but it is.
No that is not correct, as I read the problem.
If $a=1$ then $x^2-2x+1=(x-1)^2$.
But $(x-1)^2=0$ if $x=1$.
Therefore $\log(x^2 -2ax+a)$ is not defined for $x=1$; i.e. for all x.

5. Well.
I take the answer as incorrect in that case.

Thank you.