$\ln (2 x-5)>\ln (7-2 x)$ Solve
$$\ln (2 x-5)>\ln (7-2 x)$$
The answer is given as  $$3<x<7/2$$
This is what I have done $$\ln (2 x-5)-\ln (7-2 x)>0$$
$$\ln \left(\frac{2 x-5}{7-2 x}\right)>0$$
However I am not able to understand how to get to the answer provided.

Exponentiation is a one to one, increasing function so:
$\ln (2x-5) > \ln (7-2x)\\
e^{\ln (2x-5)} > e^{\ln (7-2x)}\\
2x-5>7x-2\\
...\\
x>3$
However there are the domains to consider as well:
$2x-5>0$ gives $x>\frac{5}{2}$ and $7-2x>0$ gives $x<7/2$.  We have to put all of the inequalities $x>3$ and $x>\frac{5}{2}$ and $x<\frac{7}{2}$ Together.  This gives the answer $3<x<\frac{7}{2}$
