$\log _{x+\frac{7}{2}}\left(\frac{x-7}{2 x-3}\right)^2 \geq 0$

Feasible region : $x+\frac{7}{2}>0 \Rightarrow x > -\frac{7}{2}$

And $x+\frac{7}{2} \neq 1 \Rightarrow x \neq-\frac{5}{2}$

And $\frac{x-7}{2 x-3} \neq 0 \quad$ and $2 x-3 \neq 0$

$\Downarrow$

$x \neq 7$

$x \neq \frac{3}{2}$

Taking intersection : $x \in\left(\frac{-7}{2}, \infty\right)-\left\{-\frac{5}{2}, \frac{3}{2}, 7\right\}$

Now $\log _a b \geq 0$ if $a > 1$ and $b \geq 1$

Or

$a \in(0,1)$ and $b \in(0,1)$

$\text { C-I; } \quad x+\frac{7}{2}>1 \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \geq 1$

$x > -\frac{5}{2} \quad(2 x-3)^2-(x-7)^2 \leq 0$

$(2 x-3+n-7)(2 x-3-x+7) \leq 0$

$(3 x-10)(x+4) \leq 0$

$\quad x \in\left[-4, \frac{10}{3}\right]$

$\text { Intersection : } x \in\left(\frac{-5}{2}, \frac{10}{3}\right]$

$\text { C-II } x+\frac{7}{2} \in(0,1) \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \in(0,1)$

$0 < x+\frac{7}{2} < 1 \quad \quad\left(\frac{x-7}{2 x-3}\right)^2 < 1$

$-\frac{7}{2} < x < \frac{-5}{2} \quad(x-7)^2<(2 x-3)^2$

$x \in(-\infty,-4) \cup\left(\frac{10}{3}, \infty\right)$

No common values of $x$.

Hence intersection with feasible region

We get $x \in\left(\frac{-5}{2}, \frac{10}{3}\right]-\left\{\frac{3}{2}\right\}$

Integral value of $x$ are $\{-2,-1,0,1,2,3\}$

No. of integral values $=6$