# Doubling Time

### 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?

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.

The answer to the question about Israel in the box above 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

\[ x_1 = x_0 \times \left( 1 + \frac{n}{100} \right). \tag{1}\]

Assume that after \(y\) years the population will be double, i.e.:

\[ x_k = x_0 \times \left( 1 + \frac{n}{100} \right)^y = 2x_0. \tag{2}\]

Cancelling \(x_0\) in Equation 2 we get

\[ \left( 1 + \frac{n}{100} \right)^y = 2. \tag{3}\]

We now take the natural logarithm of both sides in Equation 3, yielding:

\[ y\ln\left( 1 + \frac{n}{100} \right) = \ln(2). \tag{4}\]

Note that we took \(y\) out of the exponent and it now multiplies the logarithm on the left-hand side. Multiplying both sides in Equation 4 by 100 yields

\[ 100\,y\ln\left( 1 + \frac{n}{100} \right) = 100\ln(2) \simeq 69.3. \tag{5}\]

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:

\[ y\ln\left( 1 + \frac{n}{100} \right)^{100} = 100\ln(2). \tag{6}\]

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

\[ \exp(x) = \displaystyle\lim_{m\rightarrow \infty} \left( 1 + \frac{x}{m} \right)^{m}. \tag{7}\]

Because the number 100 is “quite big”, we will approximate the parenthesis inside the logarithm in Equation 6 with the exponential function in Equation 7, yielding

\[ y\ln\exp(n) = 100\ln(2). \tag{8}\]

The logarithm is the inverse function of the exponential, therefore

\[ y\,n = \displaystyle100\ln(2). \tag{9}\]

Finally, solving for \(y\), we have

\[ y = \displaystyle\frac{100\ln(2)}{n} \simeq \frac{70}{n}. \tag{10}\]

We have thus shown that the number of years \(y\) 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

\[ y = \displaystyle\frac{\ln(2)}{\ln(1+n/100)}\simeq 38.85. \tag{11}\]

### conclusion

👍 Very impressive rule of thumb 👍