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?
\(A = Pe^{rt}\)
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
