Questions: Suppose that 16,745 is invested at an interest rate of 6.6% per year, compounded continuously. a) Find the exponential function that describes the amount in the account after time t, in years. b) What is the balance after 1 year? 2 years? 5 years? 10 years? c) What is the doubling time? a) The exponential growth function is P(t)= (Type exponential notation with positive exponents. Do not simplify. Use integers or decimals for any numbers in the equation.)

Find the exponential function that describes the amount in the account after time \( t \), in years.

Exponential function formulation...

The exponential function for continuous compounding is given by \( A(t) = P \cdot e^{rt} \). Substituting the values \( P = 16745 \) and \( r = 0.066 \), we have:
\[ A(t) = 16745 \cdot e^{0.066t} \]

Final expression...

Thus, the exponential function is:
\[ \boxed{A(t) = 16745 \cdot e^{0.066t}} \]

What is the balance after 1 year? 2 years? 5 years? 10 years?

Calculate balance after 1 year...

For \( t = 1 \):
\[ A(1) = 16745 \cdot e^{0.066 \cdot 1} \approx 17887.46 \]

Calculate balance after 2 years...

For \( t = 2 \):
\[ A(2) = 16745 \cdot e^{0.066 \cdot 2} \approx 19107.86 \]

Calculate balance after 5 years...

For \( t = 5 \):
\[ A(5) = 16745 \cdot e^{0.066 \cdot 5} \approx 23291.76 \]

Calculate balance after 10 years...

For \( t = 10 \):
\[ A(10) = 16745 \cdot e^{0.066 \cdot 10} \approx 32398.10 \]

The balances are:
\[ \boxed{A(1) \approx 17887.46}, \quad \boxed{A(2) \approx 19107.86}, \quad \boxed{A(5) \approx 23291.76}, \quad \boxed{A(10) \approx 32398.10} \]

What is the doubling time?

Calculate doubling time...

The doubling time \( t \) can be calculated using the formula:
\[ t = \frac{\ln(2)}{r} \]
Substituting \( r = 0.066 \):
\[ t \approx \frac{\ln(2)}{0.066} \approx 10.50 \]

The doubling time is approximately:
\[ \boxed{10.50 \text{ years}} \]

The exponential function is \( \boxed{A(t) = 16745 \cdot e^{0.066t}} \).
The balances are \( \boxed{A(1) \approx 17887.46} \), \( \boxed{A(2) \approx 19107.86} \), \( \boxed{A(5) \approx 23291.76} \), \( \boxed{A(10) \approx 32398.10} \).
The doubling time is \( \boxed{10.50 \text{ years}} \).