Question about the exponential growth and decay formula

  • Thread starter FilupSmith
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[I don't know if this is in the right topic or not so I hope I'm all good]

My question is related to the exponential growth and decay formula Q=Ae^(kt).

Simply, why is the value e used as the base for the exponent?

Does it have to be e?
If so, can anybody tell me why? Thanks


~| FilupSmith |~
 
Last edited:

olivermsun

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It doesn't have to be e — it could actually be any "base."

For example, check out the first form for exponential decay given here; the base is 2 (or 1/2) because here the exponent is scaled by the half-life.

You can convert between different bases; here 2 is easy because it makes sense in this particular problem, other times e is used because it's convenient for doing further math like taking derivatives and integrals.
 
28
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I thought so, I initially noticed this when solving for t. Because e kind of disappears once you take the natural log. I guess its really for convenience?

None the less, thank you very much!


~| FilupSmith |~
 

Matterwave

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Notice that: $$x^{kt}=e^{kt\ln x}$$

So I can convert from any base x to ##e## by multiplying the constant ##k## by ##\ln x##. So it's not mandatory that we use base e. However, base e is nice basically because of the integral:

$$\int \frac{1}{x}dx=\ln(x)+C$$

When solving the differential equation, it's sort of natural to stay in base e.
 
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That helps a lot! Thanks!


~| FilupSmith |~
 

Erland

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No, it doesn't need to be ##e##. One can use any positive number except ##1## as base. In general, if ##a>0##, ##a \neq 1##: ##e^{kt}=a^{(k/\ln a)t}##. In particular, if ##a=e^k##, we get the simple form ##e^{kt}=a^t##.

The reason that the base ##e## is often used is that it simplifies differentiation: The derivative of ##e^{kt}## is ##ke^{kt}##, while the derivative of ##a^{kt}## is ##k\ln a\, a^{kt}##.

This means that ##y=Ce^{kt}## is the general solution of the differential equation ##dy/dt=ky##, which arises naturally in the study of growth and decay. With base ##a##, the solution gets the more complicated form ##y=Ca^{(k/\ln a)t}##.

So, the base ##e## a simplifies the calculations.
 

HallsofIvy

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However, [itex]a^x= e^{ln(a^x)}= e^{x ln(a)}[/itex]. In other words, it is fairly easy to convert an exponential to any base to an exponential to any other base. We typically use "e" because it is simplest- it has the property that its derivative is just itself while the derivative of any other base is a constant, other than 1, times the function.
 

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