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Homework Help: Arbitrary Constant

  1. Apr 9, 2006 #1
    How does one get from [itex]ln x = -0.03t + c'[/itex] to [itex]x = ce^{-0.03t}[/itex]

    Why isn't it [itex]x = e^{-0.03t} + c[/tex]? where [itex]c = e^{c'}[/itex]
     
    Last edited: Apr 9, 2006
  2. jcsd
  3. Apr 9, 2006 #2
    Well, it cannot be your equation for x, the + should be *

    Look at this derivation :

    [tex]ln x = -0.03t + c'[/tex]

    [tex]x = e^{-0.03t + c'}[/tex]

    [tex]x = e^{-0.03t} * e^{c'}[/tex]

    They called the [tex]e^{c'}[/tex] the constant c.

    This can be done since you this constant must be determined using some given initial conditions like at t = 0, x must be 5 or so...

    marlon

    EDIT : if you want you can just use [tex]e^{c'}[/tex] as well, it does not really matter because of the way the constant must be calculated. The result will be the same.
     
    Last edited: Apr 9, 2006
  4. Apr 9, 2006 #3
    Thanks marlon. :)
     
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