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    Okay only need reassurance about this as I have a rough idea but i'm not 100% certain

    but say you are give an equation that follows

    n= n'e^-kt

    and you are given n, n' and t when rearranging and after using logs you end up with

    log n = -kt

    If you flip the equation do you end up with

    log (1/n) = kt

    Many thanks
    Jacobdatz
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      (Original post by Jacobdatz)
      Okay only need reassurance about this as I have a rough idea but i'm not 100% certain

      but say you are give an equation that follows

      n= n'e^-kt

      and you are given n, n' and t when rearranging and after using logs you end up with

      log n = -kt

      If you flip the equation do you end up with

      log (1/n) = kt

      Many thanks
      Jacobdatz
      Yes, more or less.
      If you rearrange and log that equation you get
      log (n/n') = -kt
      Inverting gives
      log (n'/n) = kt

      The rule is that if log n = x then log (1/n) = -x
     
     
     
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