$\ln (2 x-5)>\ln (7-2 x)$ 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.

Hint 
If $$\ln \left(\frac{2 x-5}{7-2 x}\right)>0$$ it implies that $$\frac{2 x-5}{7-2 x}>1$$ But take care : the logarithm is such that its argument must be positive.
I am sure that you can take from here.
