Let f and g be functions satisfying f(x+ y) =f(x)f(y), f (l) =7 and g(x+ y) = g(xy), g(1) =1, for all $$x,y \epsilon N$$. If $$\sum_{x=1}^n \left(\frac{f(x)}{g(x)}\right) = 19607$$, then n is equal to:

Given $$f(x+y) = f(x)f(y)$$ with $$f(1) = 7$$, and $$g(x+y) = g(xy)$$ with $$g(1) = 1$$, for all $$x, y \in \mathbb{N}$$.

From $$f(x+y) = f(x)f(y)$$: $$f(n) = [f(1)]^n = 7^n$$.

From $$g(x+y) = g(xy)$$: Setting $$x = n, y = 1$$: $$g(n+1) = g(n)$$. So $$g$$ is constant for all $$n \geq 2$$: $$g(n) = g(2)$$.

Setting $$x = y = 1$$: $$g(2) = g(1) = 1$$. So $$g(n) = 1$$ for all $$n \geq 1$$.

$$\sum_{x=1}^n \frac{f(x)}{g(x)} = \sum_{x=1}^n 7^x = \frac{7(7^n - 1)}{7 - 1} = \frac{7^{n+1} - 7}{6}$$

Setting equal to 19607:

$$7^{n+1} - 7 = 6 \times 19607 = 117642$$

$$7^{n+1} = 117649$$

$$7^6 = 117649$$, so $$n + 1 = 6$$, giving $$n = 5$$.

The correct answer is Option D: 5.