Suppose that $\$ 10,227$ is invested at an interest rate of $5.3 \%$ per year, compounded continuously.
a) Find the exponential function that describes the amount in the account after time t, in years.
b) Whąt 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.)

Step 1 ：a) Recall the formula for continuous compounding: \(A = Pe^{rt}\), where \(A\) is the end balance, \(P\) is the principal, \(r\) is the interest rate, and \(t\) is the number of years.

Step 2 ：Substitute the given information: \(A = 10227e^{0.053t}\). This is the exponential function that describes the amount in the account after time \(t\), in years.

Step 3 ：b) To find the balance after 1, 2, 5, and 10 years, substitute the corresponding values of \(t\) into the function:

Step 4 ：After 1 year: \(A = 10227e^{0.053(1)}\approx 10762.97\)

Step 5 ：After 2 years: \(A = 10227e^{0.053(2)}\approx 11329.87\)

Step 6 ：After 5 years: \(A = 10227e^{0.053(5)}\approx 13716.62\)

Step 7 ：After 10 years: \(A = 10227e^{0.053(10)}\approx 18807.84\)

Step 8 ：c) To find the doubling time, set \(A = 2P\) and solve for \(t\):

Step 9 ：\(2(10227) = 10227e^{0.053t}\)

Step 10 ：\(2 = e^{0.053t}\)

Step 11 ：Take the natural logarithm of both sides: \(\ln{2} = 0.053t\)

Step 12 ：Solve for \(t\): \(t = \frac{\ln{2}}{0.053} \approx 13.08\) years

Step 13 ：The doubling time is approximately \(\boxed{13.08}\) years.