$\ln (2 x-5)>\ln (7-2 x)$ 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.
 A: 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.
A: 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 $$\therefore 3<x<\frac72$$
A: 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 $3<x<\frac{7}{2}$
