If the fourth term in the expansion of $$\left(x + x^{\log_2 x}\right)^7$$ is 4480, then the value of $$x$$ where $$x \in N$$ is equal to:

We need the fourth term in the expansion of $$\left(x + x^{\log_2 x}\right)^7$$.

Using the binomial theorem, the general term is $$T_{r+1} = \binom{7}{r} x^{7-r} \cdot \left(x^{\log_2 x}\right)^r = \binom{7}{r} x^{7-r+r\log_2 x}$$.

For the fourth term, $$r = 3$$: $$T_4 = \binom{7}{3} x^{4 + 3\log_2 x} = 35 \cdot x^{4 + 3\log_2 x}$$.

We are given $$T_4 = 4480$$, so $$35 \cdot x^{4 + 3\log_2 x} = 4480$$, which gives $$x^{4 + 3\log_2 x} = 128$$.

Let $$\log_2 x = t$$, so $$x = 2^t$$. Substituting: $$(2^t)^{4 + 3t} = 128 = 2^7$$.

This gives $$2^{t(4 + 3t)} = 2^7$$, so $$t(4 + 3t) = 7$$, i.e., $$3t^2 + 4t - 7 = 0$$.

Using the quadratic formula: $$t = \frac{-4 \pm \sqrt{16 + 84}}{6} = \frac{-4 \pm 10}{6}$$.

So $$t = 1$$ or $$t = -\frac{7}{3}$$.

Since $$x \in \mathbb{N}$$, we need $$x = 2^t$$ to be a natural number. For $$t = 1$$: $$x = 2^1 = 2$$. For $$t = -\frac{7}{3}$$: $$x = 2^{-7/3}$$, which is not a natural number.

Therefore $$x = 2$$, which matches Option A.