<s>[INST] $\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right)<1 $ $$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right) $$
Which is the solution of$\ f(x)<1 $ ? I did:
$$\ \frac{2x+3}{7-x}>0 $$
$\ x<7 $ results in $\ x> \frac{-3}{2} $ whereas $\ x>7 $ results in $\ x< \frac{-3}{2} $ so x=($\ \frac{-3}{2} $,7) and also
$\ x-2>0 $ so $\ x>2 $
$$\ x=(2,7)-\{3\} $$
So after finding out where $\ f $ is defined, I did this:
$$\log _{x-2}\left(\frac{2x+3}{7-x}\right) <\log \:_{x-2}\left(x-2\right) $$
$$\ \frac{2x+3}{7-x} <x-2 $$
$$\ x^2-7x+17<0 $$
And I figured I must have done a mistake somewhere as $\ d<0 $ and $\ x^2 $'s coefficient is $\ >0 $ so $\ f $ would be $\ >0$. Could you let me know where is my mistake? The correct answer is $\ x=(2,3) $
 [/INST] The domain gives $2<x<7$, $x\neq3$ and since $\frac{2x+3}{7-x}\neq x-2$,
we get the answer without any cases immediately by the intervals method:
$$(2,3).$$
Done!
</s>