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

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 \(=6\)