Why is e used in exponential growth functions?Now I'm aware that e x is its own derivative, which makes it convenient to use in calculus. However, I have a question about this function:Intuitively, an exponential growth function could be written as a ∗ ( 1 + k ) t , where a is the initial amount, k is the growth rate and t is the time.However, it can also be rewritten as a ∗ e k t , same thing, but with Euler's number added to it.In what way are these both the same? They yield the same answer. Why not 2, or 4, or 10? I know that one would need to change the growth constant if using one of those bases, the one I called k, but how come one doesn't need to change it when it's in base e?

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Why is e used in exponential growth functions?Now I&#039;m aware that       e    x   is its own derivative, which makes it convenient to use in calculus. However, I have a question about this function:Intuitively, an exponential growth function could be written as   a  ∗  (  1  +  k      )    t  , where a is the initial amount, k is the growth rate and t is the time.However, it can also be rewritten as   a  ∗      e          k      t      , same thing, but with Euler&#039;s number added to it.In what way are these both the same? They yield the same answer. Why not 2, or 4, or 10? I know that one would need to change the growth constant if using one of those bases, the one I called k, but how come one doesn&#039;t need to change it when it&#039;s in base e?

yen1291kp6 · Accepted Answer

The two k&#039;s in your formula will be different in order to make the two the same. Let&#039;s go with a simple concept in exponential growth:  Doubling time. As a concrete example, say we have a population that doubles every 4 units of time (Let&#039;s say days), so t is measured in days. Then it is pretty clear that the population at any given time is given by  P  (  t  )  =      P    0    ⋅      2                  t        4            where       P    0   is the initial population. In your non-e based formulation, this would be  P  (  t  )  =      P    0    ⋅  (      2                  1        4                  )    t    =      P    0    ⋅  (  1  +  (      2                  1        4              −  1  )      )    t  thus making your growth rate k in this version       2                  1        4              −  1 Not very intuitive or easy to work with.Changing it to base e just takes writing 2 as       e          ln      ⁡      2       and using exponent laws:  P  (  t  )  =      P    0    ⋅      2                  t        4              =      P    o    ⋅      e          ln      ⁡      2      ⋅              t        4              =      P    0    ⋅      e                            ln          ⁡          2                4            ⋅      t      This is the standard version of k we would use for this problem   k  =            ln      ⁡      2        4  . So you see you can write it either way, you just need to change how you define k.Using e is natural because we use these functions in differential equations all the time so using any other base would require constant change of bases. Finding k is easy, you just replace 4 with whatever the doubling time is in units of time. For exponential decay, you do the exact same but make it negative (because   ln  ⁡  (      1    2    )  =  −  ln  ⁡  2

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