Animated Solution for Mathematics - Basic Mathematics: The number of integral solution $x$ of ${\log }_{(x+\frac{7}{2})}{\left(\frac{x-7}{2x-3}\right)}^2\ge 0$ is

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving an inequality; we are embarking on a journey through the logical architecture of logarithms. The problem before us is:

It looks intimidating, but in mathematics, fear is just a lack of structure. Let us build that structure together.

Before we touch a single variable, we must respect the 'Gatekeepers'. For ${\log }_{g(x)}f(x)$ to be defined, three conditions must be satisfied simultaneously:

1. The argument $f(x)$ must be strictly positive: $f(x)>0$. 2. The base $g(x)$ must be strictly positive: $g(x)>0$. 3. The base $g(x)$ cannot be unity: $g(x) eq 1$.

For our base, $x+\frac{7}{2}>0$ implies $x>-\frac{7}{2}$. Furthermore, $x + \frac{7}{2} eq 1$ implies $x eq -\frac{5}{2}$.

The argument ${\left(\frac{x-7}{2x-3}\right)}^2$ is a perfect square, so it is always non-negative. It is zero when $x=7$ and undefined when $2x-3=0$, or $x=\frac{3}{2}$. Thus, our domain is the entire real line, excluding $\{-3.5,-2.5,1.5,7\}$.

The behavior of a logarithm depends entirely on its base. We must split our analysis into two distinct cases.

Case I: The Shrinking Base ($0<\text{base}<1$) When the base is between $0$ and $1$, the function is strictly decreasing, and the inequality sign must flip. Our base $x+\frac{7}{2}$ is between $0$ and $1$ when $-3.5<x<-2.5$.

In this region, the inequality transforms into:

Testing the only integer in this range, $x=-3$, we find the argument becomes ${\left(\frac{-10}{-9}\right)}^2\approx 1.23$. Since $1.23$ is not less than or equal to $1$, Case I yields no solutions.

Case II: The Growing Base ($\text{base}>1$) When the base is greater than $1$, the function is increasing, and the inequality sign remains unchanged. This occurs when $x>-2.5$. Our inequality becomes:

To solve this, we bring the $1$ to the left and use the difference of squares:

Subcase A: $\frac{x-7}{2x-3}\ge 1$. Simplifying leads to $\frac{x+4}{2x-3}\le 0$. Using the Wavy Curve method, the solution is $x\in [-4,1.5)$. Intersecting this with our Case II condition ($x>-2.5$), we get $x\in (-2.5,1.5)$.

Subcase B: $\frac{x-7}{2x-3}\le -1$. Simplifying leads to $\frac{3x-10}{2x-3}\le 0$. Using the Wavy Curve method, the solution is $x\in (1.5,\frac{10}{3}]$. Since this entire interval is greater than $-2.5$, it is fully valid.

We have arrived at the finish line. Our valid intervals are $(-2.5,1.5)$ and $(1.5,\frac{10}{3}]$.

Counting the integers: - In $(-2.5,1.5)$, the integers are $\{-2,-1,0,1\}$ ($4$ values). - In $(1.5,3.33]$, the integers are $\{2,3\}$ ($2$ values).

Adding them together, $4+2=6$. We have found $6$ integral solutions.