We are given that the function has the form:

$f(t) = A e^{-kt}$

where $A$ and $k$ are constants. Additionally, we know:

$f(1) = 1 \quad \text{and} \quad f(2) = 13$

Step 1: Find $A$ and $k$

We can use the given values to form a system of equations.

1. From $f(1) = 1$:

$f(1) = A e^{-k(1)} = A e^{-k} = 1$ So we have the equation: $A e^{-k} = 1$ This gives us: $A = e^{k}$

2. From $f(2) = 13$:

$f(2) = A e^{-k(2)} = A e^{-2k} = 13$ Substitute $A = e^{k}$ into this equation: $e^{k} e^{-2k} = 13$ Simplifying: $e^{-k} = 13$ So: $-k = \ln(13)$ $k = -\ln(13)$

Step 2: Find $f(3)$

Now that we have $k = -\ln(13)$ and $A = e^k = e^{-\ln(13)} = \frac{1}{13}$, we can find $f(3)$.

$f(3) = A e^{-k(3)} = \frac{1}{13} e^{-(-\ln(13))(3)} = \frac{1}{13} e^{3\ln(13)}$ Simplifying $e^{3\ln(13)}$: $e^{3\ln(13)} = 13^3$ Therefore: $f(3) = \frac{1}{13} \times 13^3 = 13^2 = 169$

Final Answer:

$f(3) = 169$

Would you like further clarification or details on any part of the solution? Here are a few follow-up questions for you:

1. What does the constant $A$ represent in the function $f(t) = Ae^{-kt}$?
2. How would the function $f(t)$ behave if $k$ were positive instead of negative?
3. What would happen to $f(t)$ if $k = 0$?
4. Can we apply this type of exponential decay function in real-world scenarios? Provide an example.
5. How do we interpret the result $f(3) = 169$ in the context of the problem?

Tip: In problems like this, exponential functions often model processes like radioactive decay, population growth, or cooling, where one quantity diminishes or increases at a rate proportional to its current value.