$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right)<1 $

Question

$$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right) $$ 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) $

Answer 1

$$\log _{(x-2)}\left(\frac{2x+3}{7-x}\right)<1$$

Hint

$1)$ $x-2>0$ and $x-2\ne 1$

$2)$ $\frac{2x+3}{7-x}>0$

$2)$ If $x-2>1$ then

$$\frac{2x+3}{7-x}<x-2$$

$3)$ If $0<x-2<1$ then

$$\frac{2x+3}{7-x}>x-2$$

Now you can solve it.

Answer 2

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!