# Why do we use base $e$ in population growth questions?

I know that we need base e to differentiate but I don't see what makes this formula work.

$$P = P_0 e^{rt}$$

where the 0 refers to initial population, $r$ the rate, and $t$ the time.

Changing the base changes the curve, so why does base e work? I mean $r$ and $t$ are pretty straightforward numbers so there's no fancy constants (other than $e$). Why is it not base $2$ or something else?

• Do you know, where this equation comes from? – Kaster Jul 6 '15 at 2:51
• Have you learned calculus yet? If not, I suspect your question will merely be reduced to "What's so great about $e$ as the base of an exponential function?" – pjs36 Jul 6 '15 at 3:18
• I know enough calculus to know that $e^x$ differentiates to itself. – Dylanthepiguy Jul 6 '15 at 7:44

Two issues: (1) Why is base $e$ used?; and (2) "Changing the base changes the curve." is wrong if you do things right.

Suppose the population doubles every $30$ years. Then what is the population after $180$ years?

Notice that $180/30=6$, i.e. $6$ is the number of $30$-year periods and thus the number of doublings. So the population will be $P_0\cdot2^6 = P_0\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2 = 64 P_0$.

What is the population after $t$ years? The number of $30$-year periods is $t/30$, so that's the number of doublings, and it is $P_0 \cdot 2^{t/30}$.

Notice that base $e$ is nowhere used above.

But what is the instantaneous rate of change at time $t=0$? It is $$\left.\frac{dP}{dt}\right|_{t=0} = P_0 \frac d {dt} 2^{t/30} = P_0 \cdot 2^{t/30} (\log_e 2)\cdot 1 {30}.$$ If you're not dealing with instantaneous rates of change, then you don't need $e$.

Suppose now we're told the initial population is $P_0$ and it's growing at $P_0/40$ per year. How long will it take to double? It follows that $$P= P_0 e^{t/40}. \tag 1$$ If $t$ is the doubling time, then $e^{t/40}=2$, so $\dfrac t {40} =\log_e 2$ and the doubling time is $t=40\log_e 2 = 27.725887\ldots\text{ years}$. Again, we need $e$ only because instantaneous rates of change are involved.

Notice that $\dfrac d {dt} 8^t = (8^t\cdot\text{constant})$. If the base had been $6$ rather than $8$, the constant would be different. Only when the base is $e$ is the constant $1$. That is what is "natural" about $e$.

Above we found that the doubling time is $27.725887\ldots$ years, and the first argument above shows that $$P = P_0 \cdot 2^{t/27.725887\ldots}. \tag 2$$

Is $(1)$ different from $(2)$? No. They're the same. Changing the base does not change the curve if $t$ is multiplied by the constant appropriate to the base in each case.

I've seen students write things like $P = P_0 e^{-(\ln 8) t}$ and then fail to do the simplification that says this is $P = P_0\left(\frac 1 8 \right)^t$. If $t=2$, then you have $P=P_0\cdot\frac 1 {64}$, but sometimes students say $\ln 8 = 2.079$ and $e^{-(2.079)\cdot 2} = 0.015638804272$, and fail to notice that that is close to $1/64$, and think that by adding lots of digits they're making it very accurate. Those later digits are garbage. Notice that $1/64 = 0.015625$ and compare that with that previous number.

$\displaystyle P = P_0 e^{-(\ln 8) t}$ is the same as $\displaystyle P = P_0\left(\frac 1 8 \right)^t$, so changing the base doesn't change the curve if things are done correctly.

A lot of the other answers danced around the answer, so I'm going to just give you a straight one, since you said you knew a little Calculus.

If $r$ is the rate of change in a population $P$, then the actual change (in terms of the number of members) is given by $rP$. In other words, $$\frac{dP}{dt} = rP$$ Now, to move from a dynamic equation of rates to a static equation (i.e., one without differentials) you have to integrate. That means you need to separate the variables so that they are together by themselves with their differentials. You can do that in this equation by dividing both sides by $P$ and multiplying both side by $dt$. This will yield the following equation: $$\frac{dP}{P} = r\,dt$$ Now, we can integrate both sides: $$\int \frac{dP}{P} = \int r\,dt$$ These are basic integral rules (remember, $r$ is a constant!). Therefore, the equation becomes: $$\ln(P) = r\,t + C$$ However, we don't want the equation for the log of the population, we want the equation for $P$ itself, therefore we can exponentiate both sides of the equation to get: $$e^{\ln(P)} = e^{r\,t + C}$$ This then reduces to: $$P = e^{r\,t + C}$$ Now, before going further, the actual answer to your question is now given. You have an $e$ because the integral gave you an $\ln(P)$, and exponentiation by $e$ is what converts $\ln(P)$ to just $P$.

However, to finish the equation, let's use exponent rules to reduce this further. We can convert the right-hand exponentiation to: $$P = e^Ce^{r\,t}$$ However, $e^C$ is just a combination of constants, so it will itself be a constant that is still unknown. So, we can substitute $e^C$ with just $C$, giving us: $$P = C\,e^{r\,t}$$ So, what is $C$? Well, sometimes you can see what something is by setting other things to zero. So, let's set $t = 0$ and see what pops out: $$P = C\,e^{r(0)} \\ P = C\,e^{0} \\ P = C\cdot 1 \\ P = C$$ So, as you can see, $C$ is just the initial population at time $t = 0$. So, we often represent the "first" of something as $P_0$. So $C = P_0$. Now the equation becomes: $$P = P_0 e^{r\,t}$$ This is the equation where you were wondering where it came from. Hopefully all of the pieces are now clear.

The $e$ comes from the fact that integrating $\frac{dP}{P}$ yield $\ln(P)$, but what we really wanted was $P$ itself, and exponentiating both sides with $e$ will yield that.

We define $e = \displaystyle \lim_{n\to \infty} \left(1+\dfrac{1}{n}\right)^n$. The reason it shows up, why we have this definition, is the following:

Say a population, $P_0$ doubles in a "unit" of time, $t'$ (we set this to $1$ for convenience right now). Then the population at a unit of time $t'$ is $P_0 *\left(1+\frac{1}{1}\right)$. Say you want to chop this up more say it up more, like twice. This means that at half a unit of time, you get half of the population to grow, and at full you get another half. The population is then $P_0 \left(1+\frac{1}{2}\right)^2$. In general, if you want to have your population growing $n$ times, the population in a unit of time is $\left(1+\frac{1}{n}\right)^n$

However, populations are ALWAYS growing, so you take the limit for continuously compounded population growth, and you get $e$.

Furthermore we can show that $e^{rt}=\lim\limits_{n\to\infty} \left(1+\frac{rt}{n}\right)^n$.

Now your question is, why can't we have a different base? Well the answer is that the $r$ takes care of it. Lets say you have a model:

$$P(t)=P_0 e^{rt}$$

and you want it in with a base $2$. Well simply note that $e^{rt}=(e^r)^t$. So if you want it in base $2$, let $r'=\log_2(e^{r})$. Then:

$$P(t)=P_0 e^{rt}=P_0 2^{\log_2(e^{r})t}=P_0 2^{r' t}$$

Note that, according to your equation,

$$\frac{\mathrm{d}P}{\mathrm{d}t} = rP,$$ and $P(t=0) = P_0$. If $t$ indicates time and $P$ population, what is $r$? What does this equation tell us? How do you integrate it in order to get your solution?

"e" has come in by solving a differential equation for continuous growth.

It is not necessarily there in a population growth equation, which could be of the form of

P = $P_0\cdot{(1+r)}^t$ or the even more general equation y = $a\cdot b^x$

The first is is of the form of compound interest, and the second uses growth factor "b" rather than growth rate (decimals) "r"

Note that $e = \displaystyle \lim_{n\to \infty} \left(1+\dfrac{1}{n}\right)^n$, and this is the continuous growth or decay problem, then you take the limit as $n$ to infinity and get $e$.