How Does Base e Apply to Different Rates of Exponential Growth?

[Atomos]

I understand why base e can be used when the percent growth per annum is 100%, but I don't understand how it can be justified that the growth for numbers other than 1 can be put into the exponent.

 

[Tide]

Consider the limit as $n \rightarrow \infty$ of $\left(1 + rt/n\right)^n$. For nonzero r and t, define $\nu = n/(rt)$. Then, the expression becomes

[tex]\left[ \left(1 + \frac {1}{\nu} \right)^\nu\right]^{rt}[/itex]

You should be able to see your way through that. :)

 

[HallsofIvy]

I understand why base e can be used when the percent growth per annum is 100%, but I don't understand how it can be justified that the growth for numbers other than 1 can be put into the exponent.

I have no idea what you mean by that! Why should "e" work when the growth per annum is 100%? That would be doubling every year wouldn't it? (And why would "per year" be important? Couldn't you take any unit of time and get the same basic formula?)

If you have 100% increase per year then it should be easy to see that the growth formula is P(t)= P₀2^t where t is measured in years. The reason why you see base e again and again is because it has an easy derivative (and anti-derivative): the derivative of e^at is ae^at.

You CAN use that in a formula because all exponentials are interchangeable:
$$2^x= e^{ln(2^x)}= e^{xln2}= e^{kx}$$
with k= ln2.

For some problems, say "half-life" problems where a substance decreases by 1/2 in time T, it might be reasonable to write
$$M= M_0\left(\frac{1}{2}\right)^{\frac{t}{T}}$$
but if you are going to be taking derivatives or anti-derivatives, it might be better to convert to M₀e^kt where
$$k= -\frac{ln(2)}{T}$$.