1. The problem gives two functions: $$g(x) = 2x+2$$ and $$f \circ g(x) = 8x+13$$. We need to find $$x$$ such that $$g \circ f(x) = 20$$. 2. First, understand the notation: $$f \circ g(x) = f(g(x))$$ and $$g \circ f(x) = g(f(x))$$. 3. We know $$f(g(x)) = 8x + 13$$, and since $$g(x) = 2x+2$$, substitute inside: $$f(2x+2) = 8x + 13$$. 4. Let $$t = 2x + 2$$, then $$f(t) = 8x + 13$$, where $$x = \frac{t-2}{2}$$. 5. Rewrite $$f(t)$$ in terms of $$t$$: $$f(t) = 8 \times \frac{t-2}{2} + 13 = 4(t-2) + 13 = 4t - 8 + 13 = 4t + 5$$. 6. So the function $$f(x) = 4x + 5$$. 7. Now find $$g \circ f(x) = g(f(x)) = g(4x + 5) = 2(4x + 5) + 2 = 8x + 10 + 2 = 8x + 12$$. 8. Set $$g \circ f(x) = 20$$: $$8x + 12 = 20$$. 9. Solve for $$x$$: $$8x = 20 - 12 = 8$$ $$x = \frac{8}{8} = 1$$. Final answer: $$x = 1$$.