Let $$f(x) = 2^{10}x + 1$$ and $$g(x) = 3^{10}x - 1$$. If $$(fog)(x) = x$$, then $$x$$ is equal to:

We are given two linear functions.

First function: $$f(x)=2^{10}x+1.$$ Second function: $$g(x)=3^{10}x-1.$$

The statement $$(fog)(x)=x$$ means composition; that is, we must substitute $$g(x)$$ into $$f(x)$$ and equate the result to $$x$$. By definition of composition,

$$ (fog)(x)=f\!\left(g(x)\right). $$

Now we actually perform this substitution. Wherever we see an $$x$$ in $$f(x)=2^{10}x+1$$ we replace it by $$g(x)=3^{10}x-1$$:

$$ f\!\left(g(x)\right)=2^{10}\bigl(g(x)\bigr)+1 =2^{10}\bigl(3^{10}x-1\bigr)+1. $$

The condition tells us that this must equal $$x$$, hence

$$ 2^{10}\bigl(3^{10}x-1\bigr)+1 = x. $$

We expand the left-hand side:

$$ 2^{10}\cdot3^{10}x - 2^{10} + 1 = x. $$

Collect all terms on one side so that everything involving $$x$$ is together:

$$ 2^{10}3^{10}x - x - 2^{10} + 1 = 0. $$

Factor out $$x$$ from the first two terms:

$$ x\bigl(2^{10}3^{10}-1\bigr) -\bigl(2^{10}-1\bigr)=0. $$

Now isolate $$x$$ by adding $$2^{10}-1$$ to both sides and then dividing:

$$ x\bigl(2^{10}3^{10}-1\bigr)=2^{10}-1, $$

so

$$ x=\frac{2^{10}-1}{2^{10}3^{10}-1}. $$

The fraction still does not look exactly like any option because the options contain negative exponents. We therefore manipulate our expression to reveal that form.

Notice that multiplying the numerator and denominator by $$2^{-10}$$ will introduce the reciprocals required. Explicitly:

$$ x=\frac{2^{10}-1}{2^{10}3^{10}-1}\times\frac{2^{-10}}{2^{-10}} =\frac{(2^{10}-1)2^{-10}}{(2^{10}3^{10}-1)2^{-10}}. $$

Compute each product:

Numerator:

$$ (2^{10}-1)2^{-10}=2^{10}2^{-10}-1\cdot2^{-10}=1-2^{-10}. $$

Denominator:

$$ (2^{10}3^{10}-1)2^{-10}=3^{10}(2^{10}2^{-10})-1\cdot2^{-10}=3^{10}-2^{-10}. $$

Therefore

$$ x=\frac{1-2^{-10}}{3^{10}-2^{-10}}. $$

This is exactly the expression listed in Option B.

Hence, the correct answer is Option B.