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

Which is the solution of $\text{}f(x)1$ ? I did:

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

$\text{}x7$ results in $\text{}x\frac{-3}{2}$ whereas $\text{}x7$ results in $\text{}x\frac{-3}{2}$ so x=($\text{}\frac{-3}{2}$,7) and also $\text{}x-20$ so $\text{}x2$

$$\text{}x=(2,7)-\{3\}$$

So after finding out where f is defined, I did this:

$${\mathrm{log}}_{x-2}\left(\frac{2x+3}{7-x}\right)<\mathrm{log}{\phantom{\rule{mediummathspace}{0ex}}}_{x-2}(x-2)$$

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

$$\text{}{x}^{2}-7x+170$$

And I figured I must have done a mistake somewhere as $\text{}d0$ and $\text{}{x}^{2}$'s coefficient is $\text{}0$ so $\text{}f$ would be $\text{}0$. Could you let me know where is my mistake? The correct answer is $\text{}x=(2,3)$

Which is the solution of $\text{}f(x)1$ ? I did:

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

$\text{}x7$ results in $\text{}x\frac{-3}{2}$ whereas $\text{}x7$ results in $\text{}x\frac{-3}{2}$ so x=($\text{}\frac{-3}{2}$,7) and also $\text{}x-20$ so $\text{}x2$

$$\text{}x=(2,7)-\{3\}$$

So after finding out where f is defined, I did this:

$${\mathrm{log}}_{x-2}\left(\frac{2x+3}{7-x}\right)<\mathrm{log}{\phantom{\rule{mediummathspace}{0ex}}}_{x-2}(x-2)$$

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

$$\text{}{x}^{2}-7x+170$$

And I figured I must have done a mistake somewhere as $\text{}d0$ and $\text{}{x}^{2}$'s coefficient is $\text{}0$ so $\text{}f$ would be $\text{}0$. Could you let me know where is my mistake? The correct answer is $\text{}x=(2,3)$