Suppose that$10,055 is invested at an interest rate of 5.6% 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?

Guest Feb 26, 2017

#1**0 **

\(A = Pe^{rt}\)

e is about equal to 2.71828182846...

a) \(A = 10,055e^{0.056*year}\)

b)

1 Year = $10634.14

2 Years = $11246.64

5 Years = $13304.07

10 Years = $17603.01

c) \(10055*2=10055e^{0.056t}\)

\(2=e^{0.056t}\)divide both sides by 10055

\(ln 2 = 0.056t\)take the natural log of both sides

\(ln2/0.056 = t\)divide both sides by 0.056

\(ln 2 = log_e 2\)ln is just natural log, which is log base e

t = 12.3776282242847375

Guest Feb 27, 2017