First the argument of a logarithm function must be positive.
insures that is true for both sides.
Second, the logarithm function is injective (one-to-one) on its domain.
Therefore solve .
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.