Using Monotonicity to find the value of a required variable Given $f(x) = \log_c\frac{x-2}{x+2}$, is defined $\forall \; x \in [a,b]$ and the function is monotonically decreasing.
We have to find the value (or range of values) of $c$ such that there exists $a$ and $b$, where $(2 < a < b)$, and the range of the function over $[a,b]$ is $[\log_cc(b-1), \log_cc(a-1)]$.
I have concluded the following:

*

*$c \in (0,1)$

*Since the function is monotonically decreasing $$f(a) = \log_cc(a-1) \implies \frac{a-2}{a+2} = c(a-1)$$ and $$f(b) = \log_cc(b-1) \implies \frac{b-2}{b+2} = c(b-1)$$
I thought that subtracting the two equations or taking their ratio might help me find some helpful inequality, but unfortunately couldn't find anything.
Your help is appreciated. Thanks
UPDATE $1$: We know that $c \in (0,1)$ and $\frac{x-2}{x+2} \in (0,1) \; \forall \; x \in (2, \infty)$. Hence, the value attained by the function is positive. Thus,
$$\log_cc(a-1) > 0 \implies c(a-1) < 1 \implies a < \frac{1}{c} + 1$$
and a similar inequality for $b$.

As you wrote, we have
$$\frac{a-2}{a+2} = c(a-1)$$
which is equivalent to
$$ca^2+(c-1)a-2c+2=0\tag1$$
Similarly, we have
$$cb^2+(c-1)b-2c+2=0\tag2$$
From $(1)(2)$, we want to find $c\ (0\lt c\lt 1)$ such that the equation
$$cx^2+(c-1)x-2c+2=0$$
has two distinct real roots $x$ satisfying $x\gt 2$.
So, we want to find $c$ satisfying
$$\begin{cases}0\lt c\lt 1
\\\\(c-1)^2-4c(-2c+2)\gt 0
\\\\\dfrac{1-c}{2c}\gt 2
\\\\c\cdot 2^2+(c-1)\times 2-2c+2\gt 0\end{cases}$$
So, the answer is $0\lt c\lt \dfrac 19$.
