# Exponential growth: deriving the growth factor

Basic exponential growth is

$$x(t) =ab^{t/\tau}$$

where $b$ is the growth rate or factor. Now, as Wikipedia describes with a bacteria example, the growth rate is $2$. This comes from starting with one bacteria (initial condition), then "doubling," which means $100\%$, or 1.0, i.e., $1 + 1.0$ means the growth rate is $2$.

This is very confusing to a beginner like me. Investigation led me to this alternate formula for exponential growth, which seems to break down the growth rate:

$$y = a(1 + r)^x$$

that is, $(1 + r)$ is $b$, although I'm guessing the initial starting point has only one single item. So, e.g., if there were two initial bacteria, we would have

$$y = a(2 + r)^x$$

Is this true? Basically, I'd like to see how we can say $(1 + r)$ is derived from just plain old $b$. I'm not finding a source that explains this very well. The Wikipedia certainly doesn't.

The form $ab^{t/\tau}$ has more free parameters than it really needs. You can decide, once and for all, what your base $b$ is, and then it is not part of the model anymore. Or you can decide, once and for all, what the coefficient $1/\tau$ in the exponent is, and then it is not a part of the model anymore. People use both approaches. Generally if they fix the base, they fix it to be either $2$ (for "doubling" or "halving" problems), $e$ (for calculus convenience), or sometimes $(1+r)$ (usually just for compound interest). Generally if they fix the coefficient, they fix it to be $1$. In all cases, this decision is a matter of convenience, because an exponential model is really determined by two parameters, not three.
The initial quantity in an exponential problem appears as the coefficient on the outside; for example if a population of bacteria double every $\tau$ time units, and they start out at a population of $10$, then their population can be modeled as $10 \cdot 2^{t/\tau}$.
• I thought $(1 + r)$ in the bacteria meant we start with one bacteria, and then double it, i.e., 2% growth in decimal notation is $0.02$ and 100% growth in decimal notation is therefore $1.0$; hence, $1 + 1.0 = 2$. But now I'm confused. If we start with three bacteria, then $3 + 1.0 = 4$, which is bogus. So, e.g., 2% growth is simply $1.02$, then times the initial population $P_0$ to give the next round of $P_1$. That means 100% is simply $2.0$. But I'd still like to see some proof, or something deeper explaining the seemingly arbitrary way we got 100% means $2$ in this formula. – 147pm Oct 10 '17 at 3:10
• @147pm As I said, the starting amount is on the outside, not in the base. So if you have 3 bacteria growing by 2% each time unit then that's described by $3 \cdot 1.02^t$. And yes, an increase of 100% is 2.0...simply because it means you add the starting amount to itself, so you're left with twice that. – Ian Oct 10 '17 at 4:11
• Yes, this is simply one of those arithmetic tricks, such as figuring out how to add tax, i.e., multiply by $1.09$ if the tax is $9\%$. There is no "proof" per se, it's just a trick. Still, Wikipedia doesn't explain this, which is confusing for beginners. So I'm someone who "gets to the bottom" of this issue. How many others simply give up? Alas. . . . – 147pm Oct 10 '17 at 12:15
The point is that the number of new bacteria in a period is proportional to the number at the start of the period. If $1$ bacterium becomes $1+r$, two bacteria will become $2(1+r)=2+2r$. Thus we can replace $1+r$ by $b$. The initial quantity can be factored out and becomes a multiplier.
• Yes, agreed. One text called the increase per round a "constant ratio," e.g., a $2^x$ sequence increases per round by doubling the previous number of the sequence. Thanks. – 147pm Oct 10 '17 at 12:18