4) \( { }^{5 x} \), Both of these functions grow as \( x \) gets larger and larger. Which function eventually exceeds the other? \( f(x)=8^{x-3} g(x)=10 x \) Submit

It seems that the equation \( 8^x = 5120x \) cannot be solved directly using the tools available. Instead, we can analyze the functions graphically or use a numerical approximation method. ### Step 6: Graphical Analysis To determine where \( f(x) \) exceeds \( g(x) \), we can evaluate both functions at various values of \( x \): - For small values of \( x \): - \( x = 0 \): - \( f(0) = 8^{-3} = \frac{1}{512} \) - \( g(0) = 0 \) - \( x = 1 \): - \( f(1) = 8^{-2} = \frac{1}{64} \) - \( g(1) = 10 \) - \( x = 2 \): - \( f(2) = 8^{-1} = \frac{1}{8} \) - \( g(2) = 20 \) - \( x = 3 \): - \( f(3) = 8^{0} = 1 \) - \( g(3) = 30 \) - \( x = 4 \): - \( f(4) = 8^{1} = 8 \) - \( g(4) = 40 \) - \( x = 5 \): - \( f(5) = 8^{2} = 64 \) - \( g(5) = 50 \) - \( x = 6 \): - \( f(6) = 8^{3} = 512 \) - \( g(6) = 60 \) ### Step 7: Conclusion From the evaluations, we can see that \( f(x) \) starts to grow significantly faster than \( g(x) \) as \( x \) increases. To summarize, while \( g(x) = 10x \) is larger for small values of \( x \), \( f(x) = 8^{x-3} \) will eventually exceed \( g(x) \) as \( x \) becomes large enough. Thus, the function \( f(x) \) will eventually exceed \( g(x) \).