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Laws of logs question

  1. Apr 23, 2012 #1
    I'm reading about how the chemistry kinetics equations are derived and heres something I don't get. How does this:
    http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1-6.gif
    get turned into this:
    [itex][A] = [A]_{0}*e^{-kt}[/itex]?
    When I try to derive it, I first get this:
    [itex]ln[A] - ln[A]_{0} = -kt[/itex].
    Then I isolate ln[A] and get:
    [itex]ln[A] = ln[A]_{0} - kt[/itex]
    then I reverse the ln on both sides of the equation and get:
    [itex][A] = [A]_{0} - e^{-kt}[/itex].
    I don't understand how the two terms end up multiplied rather than subtracted.
     
  2. jcsd
  3. Apr 23, 2012 #2
    The red part is the wrong part.
    By reverse you use e as a index for exponentiation, so lets say :
    ln a = ln b - c
    e^(ln a) = e^(ln b - c)
    u got your exponentialtion wrong.
     
  4. Apr 23, 2012 #3
    The easiest method to see this is to apply the exponential to each side.

    ##
    \begin{eqnarray*}
    \displaystyle ln\frac{[A]}{[A]_0} &=& -kt\\
    \displaystyle e^{ln\frac{[A]}{[A]_0}}&=& e^{-kt}\\
    \displaystyle \frac{[A]}{[A]_0} &=& e^{-kt}\\
    \displaystyle [A] &=& [A]_0 e^{-kt}\\
    \end{eqnarray*}
    ##
     
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