Sign change - Polynomial Inequality

$$f(x) = \frac{x^2 + 2x - 15}{x^2 - 7x - 18}$$<0

$$\frac{\left(x+5\right)\left(x-3\right)}{\left(x-9\right)\left(x+2\right)}<0$$

We have four inflection points -5, -2, 3, and 9.

For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive.

When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again.

We are left with the range (-5,-2) and (3,9) where the expression will be negative.

Let $$\ \log_2n=y$$

$$\ \ \dfrac{\ 4-y}{3-\dfrac{y}{2}}<0$$

$$\ \ \left(4-y\right)\left(3-\dfrac{y}{2}\right)<0$$

$$\ \ \left(4-y\right)\left(6-y\right)<0$$

$$\ \ \left(y-4\right)\left(y-6\right)<0$$

$$4 < y < 6$$

$$4<\log_2n<6$$

$$2^4<n<2^6$$

$$16<n<64$$

n can take values from 17 to 63(inclusive).

The number of n values possible = 47

There are two critical points for the inequality to consider: x=-5 and x=3/2

Region I: x is greater than 3/2
In this scenario, both the terms would be positive; cross-multiplying, we get the relation $$2x-3\le x+5$$
Giving the boundary $$x\le8$$, hence giving us the valid range as $$\frac{3}{2}<x\le8$$

Region II: $$-5<x<\frac{3}{2}$$
In this case, the right-hand side will be a negative value, and hence, the sign would change when multiplying, giving the inequality
$$2x-3\ge x+5$$
Which will give x>8, which is out of bounds for this region 

Another way is to put a value in the region to check for the validity of the inequality; by putting x=0, we could see that the inequality does not hold in this region 

Region III: x less than -5
In this scenario, both the terms are negative, essentially giving us the same boundary as region 1; we take the lower bounds, giving us that x has to be less than 5

Therefore, for the given inequality to hold true x<-5 or $$\frac{3}{2}<x\le8$$

Hence, Option A is the correct answer.