If the third term in the binomial expansion of (1+xlog₂x)⁵ equals 2560, then a possible value of x is?

(1+xlog₂x)⁵
T₃ = ⁵C₂(xlog₂x)² = 2560
→10(xlog₂x)² = 2560
→(xlog₂x)² = 256
→x² (log₂x)² = 256
Taking square root on both sides:
x log₂x = ±16
If x log₂x = 16
Let's try x = 4:
4 log₂4 = 4(2) = 8 ≠ 16
Let's try x = 16:
16 log₂16 = 16(4) = 64 ≠ 16
Let's try x = 8:
8 log₂8 = 8(3) = 24 ≠ 16
Let's try x = 2:
2 log₂2 = 2(1) = 2 ≠ 16
Let's try x = 4√2
4√2 log₂(4√2) = 4√2(2 + 1/2) = 4√2(5/2) = 10√2 ≠ 16
If x log₂x = -16
Let's try x = 1/4
(1/4) log₂(1/4) = (1/4)(-2) = -1/2 ≠ -16
Let's try x = 1/8
(1/8) log₂(1/8) = (1/8)(-3) = -3/8 ≠ -16
Let's try x = 1/16
(1/16) log₂(1/16) = (1/16)(-4) = -1/4 ≠ -16
Let's solve x log₂x = ±16 numerically or graphically.
Solving numerically or graphically we find that x = 4 or x = 1/4 are possible values. However, there seems to be a problem in the original equation given, because there is an error in the calculation. The correct solution should involve solving the equation x log₂x = ±16, which does not yield 4 or 1/4 as simple solutions.
The provided solution is flawed. A more accurate approach involves numerical methods or graphing to solve for x in the equation x log₂x = ±16.