Let's evaluate each function at the given value.

1. For the first function: $f(x) = \frac{1}{3} \cdot 6^x \quad \text{at} \quad x = 2$ Substituting $x = 2$ into the equation: $f(2) = \frac{1}{3} \cdot 6^2 = \frac{1}{3} \cdot 36 = \frac{36}{3} = 12$ So, $f(2) = 12$.

2. For the second function: $f(n) = 10 \cdot 2^n \quad \text{at} \quad n = 5$ Substituting $n = 5$ into the equation: $f(5) = 10 \cdot 2^5 = 10 \cdot 32 = 320$ So, $f(5) = 320$.

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Related Questions:

1. What happens to the first function when $x = 0$?
2. How does the second function behave for negative values of $n$?
3. Can we express the first function in terms of base 10 instead of base 6?
4. How does the second function compare to an exponential function with base 3?
5. What would happen if we change the constant multiplier in both functions?

Tip:

When dealing with exponential functions, remember that the base dictates the growth rate—higher bases lead to faster growth.