# Doubling Time

One rule of thumb to rule them all

Suppose you have a process that can be described by exponential growth. It could be anything: interests on an investment, the early phases of infection in a pandemic, whatever.

It is often convenient to have an idea how fast is the growth by answering the question:

How long will it take for $x$ to double in size, given a growth of $n$% per year?

The rule of thumb I learned a while back is the following:

Doubling time = $\displaystyle\frac{70}{n}$ (in years)

Of course, the time unit could be anything you like, I’ll deal here with years for simplicity’s sake. Specifically, let’s answer the question:

Israel has currently (2021) a population of 9.2 million, and a growth rate of 1.8% per year. How long will it take for the population to double, assuming a fixed growth rate?

The answer is about 39 years (70 divided by 1.8), but why?!

Let’s call $x_0$ the population size now, and the growth rate $n$%. After one year, the population will be

$ \displaystyle x_1 = x_0 * \left( 1 + \frac{n}{100} \right) $

Assume that after $k$ years the population will be double, i.e.:

$ \displaystyle x_k = x_0 * \left( 1 + \frac{n}{100} \right)^k = 2x_0. $

Cancelling $x_0$ we get

$ \displaystyle\left( 1 + \frac{n}{100} \right)^k = 2. $

We now take the natural logarith of both sides:

$ \displaystyle k\ln\left( 1 + \frac{n}{100} \right) = \ln(2). $

Note that we took $k$ out of the exponent and it now multiplies the logarithm on the left-hand side. Multiplying both sides by 100 yields

$ \displaystyle100k\ln\left( 1 + \frac{n}{100} \right) = 100\ln(2) \simeq 69.3. $

That surely explains the number 70 in the rule of thumb! Because of the properties of logarithms, we put the number 100 as the exponent of the parenthesis:

$ \displaystyle k\ln\left( 1 + \frac{n}{100} \right)^{100} = 100\ln(2). $

We are very close to the end! We now remind ourselves that we learned in Calculus the definition of the exponential function:

$ \exp(x) = \displaystyle\lim_{m\rightarrow \infty} \left( 1 + \frac{x}{m} \right)^{m}. $

Because the number 100 is “quite big”, we will approximate the parenthesis inside the logarithm with the exponential function, thus

$ k\ln\exp(n) = 100\ln(2). $

The logarithm is the inverse function of the exponential, therefore

$ kn = \displaystyle100\ln(2). $

Finally, solving for $k$, we have

$ k = \displaystyle\frac{100\ln(2)}{n} \simeq \frac{70}{n}. $

We have thus shown that the number of years $k$ it will take for Israel to double it’s population is about $70/n = 70/1.8 = 38.88$ years!!

The exact number, without any approximations, would be

$ k = \displaystyle\frac{\ln(2)}{\ln(1+n/100)}\simeq 38.85. $

**Conclusion:**

👍 Very impressive rule of thumb 👍