+0  
 
+5
45
1
avatar

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
Sort: 

1+0 Answers

 #1
avatar
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

6 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details