# Question about the exponential growth and decay formula

#### FilupSmith

[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

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.

#### FilupSmith

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

Gold Member
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.

#### FilupSmith

That helps a lot! Thanks!

~| FilupSmith |~

#### Erland

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

However, $a^x= e^{ln(a^x)}= e^{x ln(a)}$. 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.