To find the value of k in the given problem, we need to understand the concept of binomial expansion and its terms.

Binomial Expansion:
The binomial expansion is a way to expand a binomial expression raised to a power using Pascal's triangle and the binomial coefficients. The general form of a binomial expansion is given by:

(a + b)^n = C(n, 0)a^n b^0 + C(n, 1)a^(n-1) b^1 + C(n, 2)a^(n-2) b^2 + ... + C(n, n-1)a^1 b^(n-1) + C(n, n)a^0 b^n

Here, C(n, r) represents the binomial coefficient, which is calculated using the formula:

C(n, r) = n! / (r!(n-r)!)

In the given problem, the binomial expression is (1 + x log2x)^5.

Finding the Third Term:
To find the third term in the expansion, we can use the formula for the rth term in the binomial expansion:

T(r) = C(n, r)a^(n-r) b^r

In this case, the third term is T(3), and we can write it as:

T(3) = C(5, 3)(1)^(5-3) (x log2x)^3

Simplifying this expression:

T(3) = C(5, 3) (x log2x)^3

Given that T(3) = 2560, we can set up the equation:

C(5, 3) (x log2x)^3 = 2560

Solving for x:
To find the value of x, we need to solve the equation:

C(5, 3) (x log2x)^3 = 2560

Simplifying further:

10 (x log2x)^3 = 2560

(x log2x)^3 = 256

Taking the cube root of both sides:

x log2x = 6

x log2x can be written as x log(x)/log(2). Therefore:

x (log(x)/log(2)) = 6

x log(x) = 6 log(2)

Solving for x numerically, we find that x ≈ 1.386.

Finding the Fractional Possible Value of x:
The given problem asks for the fractional possible value of x, which can be expressed as k/4. To find k, we can multiply both sides of the equation by 4:

4x log(x) = 24 log(2)

Since x ≈ 1.386, we can substitute this value:

4(1.386) log(1.386) = 24 log(2)

Simplifying this expression using logarithmic properties:

5.544 log(1.386) = 24 log(2)

Dividing both sides by log(1.386):

5.544 ≈ 24 log(2) / log(1.386)

Calculating the right-hand side:

5.544 ≈ 24 (0.6931) / (0.1398)

5.544 ≈ 24 (4.956)

5.544 ≈ 118.944

Therefore, k = 118.