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

Good job, you're almost there:
$$\ln \left(\frac{2 x-5}{7-2 x}\right)>0 \\ \frac{2 x-5}{7-2 x}>\underbrace{e^0}_{1} \\ 2 x-5>7-2x \\ 4x>12 \\ x>3$$
That takes care of the first part, but also recall that the parameter of $\ln$ has to be greater than $0$ ($e$ to any power will not give you a number less than or equal to $0$). In other words, you also have to say this:
$$\begin{cases}2 x-5>0\\7-2 x>0\end{cases}$$
Solved:
$$\begin{cases}x>\frac52\\x<\frac72\end{cases}$$
$x$ is indeed greater than $\frac{5}{2}$, as we found out, but it must also be geater than $3$. Hence we can ignore that because we have $x>3$. Now we have an upper bound, of $x<\frac72$, because after that point the answer won't be defined. So the correct answer is $$\therefore 3<x<\frac72$$
