Finding range of a log function with variable base How can I find the range of $\log_{x-7}(x-5)$?
Here is my attempt.
Let $$y=\log_{x-7}(x-5)$$
Let $z=x-7$, $y=\log_z(z+2)$
then $y$ satisfies
$$z^y-z-2=0, z>0, z \ne 1.$$
$y$ be written as $\frac{\ln (x-5)}{\ln (x-7)}$.

The domain of the function is $x>7$ and $x \ne 8$. We note that $\ln (x-5)>0$.  $\ln(x-7)$ takes negative value when $x<8$ and it takes positive value when $x>8$.
Let's differentiate it,
$$f'(x) = \frac{\frac{\ln (x-7)}{x-5}-\frac{\ln (x-5)}{x-7}}{(\ln (x-7))^2}=\frac{(x-7)\ln (x-7) - (x-5) \ln (x-5)}{(x-5)(x-7)(\ln (x-7))^2}<0$$
Now let's examine $\lim_{x \to \infty}\frac{\ln (x-5)}{\ln(x-7)}=\lim_{x \to \infty} \frac{x-7}{x-5}=1$. Furthermore, it is clear that $1$ can't be attained.
$$\lim_{x\to 8^+}\frac{\ln (x-5)}{\ln (x-7)}=\infty$$
$$\lim_{x\to 8^-}\frac{\ln (x-5)}{\ln (x-7)}=-\infty$$
$$\lim_{x \to 7^+}\frac{\ln (x-5)}{\ln (x-7)}=0$$
Hence the range is $(-\infty, 0) \cup (1, \infty)$.
