Results 1 to 2 of 2

Math Help - proportional growth - interest rate

  1. #1
    DBA
    DBA is offline
    Member
    Joined
    May 2009
    Posts
    129

    proportional growth - interest rate

    How long will it take an investment to double in value if the interest rate is 6% compounded continuously?

    I did:

    A = A(0) * e^rt

    r=0.06
    Set A(0)=1000
    Set A=2000

    2000 = 1000 * e^0.06 * t
    ln2 = ln e^0.06 * t
    ln2 = 0.06*t
    t= ln2/0.06 = 11.55 years

    What is the equivalent annual interest rate?

    I did
    A(t)= A(0)*(1+ r/n)^nt

    2000 = 1000 (1 + r/1)^ 1*11.55
    2 = (1 + r/1)^ 1*11.55

    Here I stuck. Can someone let me know if it is right what I did and how to get the interest rate?

    I thought I could do
    ln2 = ln(1 + r/1)^ 1*11.55
    ln2 = 11.55 * ln(1 + r/1)

    but the I stuck with that...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2009
    Posts
    69
    Compounded continuously, your investment will grow according to:
     A(t) = A(0) e^{rt}
    Compounded annually, your investment will grow according to:
     A(t) = A(0) (1+r)^t ,
    where t is measured in years. The former is really just  A(t) = A(0) (e^r)^t , which is in the same form as the latter but with e^r in place of 1+r. So the equivalent rate if compounded annually is e^r - 1, which comes out to something like 6.185%.

    I think you get the same thing if you solve the equation you wrote down. You don't need to take logs to solve it, however:
     2 = (1 + r)^{11.55}
     2^{1/11.55} = 1 + r
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: September 2nd 2011, 08:26 PM
  2. Calculate interest rate based on interest in parcels
    Posted in the Business Math Forum
    Replies: 1
    Last Post: February 17th 2011, 10:38 PM
  3. Melting snowball at rate proportional to surface area?
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: July 15th 2010, 11:46 PM
  4. Replies: 2
    Last Post: October 15th 2009, 07:06 AM
  5. Replies: 2
    Last Post: April 30th 2006, 10:58 AM

Search Tags


/mathhelpforum @mathhelpforum