f(x)=log_(x-2)((2x+3)/(7-x))<1

Which is the solution of ? I did:

results in whereas results in so x=(,7) and also so

So after finding out where f is defined, I did this:
${\mathrm{log}}_{x-2}\left(\frac{2x+3}{7-x}\right)<\mathrm{log}{\phantom{\rule{mediummathspace}{0ex}}}_{x-2}\left(x-2\right)$

And I figured I must have done a mistake somewhere as and 's coefficient is so would be . Could you let me know where is my mistake? The correct answer is
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Emaidedip6g
${\mathrm{log}}_{\left(x-2\right)}\left(\frac{2x+3}{7-x}\right)<1$
Hint
$1\right)$$x-2>0$ and $x-2\ne 1$
$2\right)$ $\frac{2x+3}{7-x}>0$
$2\right)$ If $x-2>1$ then
$\frac{2x+3}{7-x}
$3\right)$ If $0 then
$\frac{2x+3}{7-x}>x-2$
Now you can solve it.
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tarjetaroja2t
The domain gives $2, $x\ne 3$ and since $\frac{2x+3}{7-x}\ne x-2$
we get the answer without any cases immediately by the intervals method:
$\left(2,3\right).$
Done!