Question

Suppose that $18,907 is invested at an interest rate of 6.5% 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.)

Answer 1

The doubling time is approximately 10.64 years.

Step-by-step explanation:

To find the exponential function that describes the amount in the account after time t, we can use the formula A = P * e^(rt), where A is the amount in the account after time t, P is the principal amount, e is Euler's number (approximately 2.71828), r is the interest rate, and t is the time in years.

In this case, the principal amount is $18,907 and the interest rate is 6.5% (or 0.065 as a decimal). So the exponential function is:

A(t) = 18907 * e^(0.065t)

For part b, we can substitute the given time values into the exponential function to find the balance. After 1 year:

A(1) = 18907 * e^(0.065*1) ≈ $20,167.56

After 2 years:

A(2) = 18907 * e^(0.065*2) ≈ $21,588.52

After 5 years:

A(5) = 18907 * e^(0.065*5) ≈ $26,855.44

After 10 years:

A(10) = 18907 * e^(0.065*10) ≈ $43,523.53

For part c, the doubling time can be found using the formula t = ln(2) / r. Substituting the value of r (0.065) into the formula:

t = ln(2) / 0.065 ≈ 10.64 years