Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

?
Lv 7
? asked in Science & MathematicsMathematics · 2 weeks ago

The solution an algebraic equation.?

Find the time t. A=P(1+r/n)^(nt), where

A=the amount after nt

P=the initial deposit

r=the annual rate of interest

n=the number of intervals in a year & n->infinity

when A=2P? Help please.

4 Answers

Relevance
  • ?
    Lv 7
    2 weeks ago
    Favorite Answer

    It would be useful to memorize the following:

    lim (1 + 1/x)^x = e

    x→∞ 

    let x = n/r

    (1+r/n)^(nt) = (1 + 1/x)^(xrt)

    = ((1 + 1/x)^(x))^(rt)

    lim ((1 + 1/x)^(x))^(rt) = e^(rt)

    x→∞

    Thus

    2P = Pe^(rt)

    t = ln(2)/r

    Because ln(2) is approximately 70% (and less-frequent compounding increases t),  this leads to the "Rule of 72": for a given rate of return r%, your investment will double in approximately 70/r or 72/r years.

  • ?
    Lv 7
    2 weeks ago

    A=P(1+r/n)^(nt)

    =>

    ln(A)=ln(P)+ntln(1+r/n)

    =>

    ln(A/P)=ln(1+r/n)/[1/(nt)]

    =>

    ln(A/P)->[1/(1+r/n)](-r/n^2)/[(-1/n^2)/t]

    as n->infinity

    =>

    ln(A/P)->rt

    =>

    A=Pe^(rt)

    When A=2P

    2P=Pe^(rt)

    =>

    t=ln(2)/r

    [Note that dependence of memorizing

    formula is not a goodway because there

    are a lot , a lot of formula in math! It  is

    a good habit to derive formula as much

    as possible when needed. Of course,

    some basic formulas are needed to

    remember as the basis]

  • 2 weeks ago

    I'm not sure what you are asking.

    If you are saying to solve for t from:

    A = P(1 + r/n)^(nt)

    and A = 2P, then we can do the substitution first, then divide both sides by P:

    2P = P(1 + r/n)^(nt)

    2 = (1 + r/n)^(nt)

    Now get the log of both sides:

    ln(2) = ln[(1 + r/n)^(nt)]

    we can pull the exponent out of the log:

    ln(2) = nt ln(1 + r/n)

    And divide both sides by [n ln(1 + r/n)]:

    ln(2) / [n ln(1 + r/n)] = t

  • 2 weeks ago

     A=P(1+r/n)^(nt)

    2P = P(1 + r/n)^(nt)

    (1 + r/n)^(nt) = 2P/P

    (1 + r/n)^(nt) = 2

    ntln(1 + r/n) = ln(2)

    ...........ln(2)

    t = ------------------

    ........nln(1 + r/n)

    ........... ln(2)

    t = ----------------------- Answer//

    .......... nln[(n + r)/n]

Still have questions? Get your answers by asking now.