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

x1=x0∗(1+n100)(1)x_1 = x_0 * \left( 1 + \frac{n}{100} \right) \tag{1}

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

xk=x0∗(1+n100)k=2x0.(2)x_k = x_0 * \left( 1 + \frac{n}{100} \right)^k = 2x_0. \tag{2}

Cancelling $x_0$ we get

(1+n100)k=2.(3)\left( 1 + \frac{n}{100} \right)^k = 2. \tag{3}

We now take the natural logarith of both sides:

kln⁡(1+n100)=ln⁡(2).(4)k\ln\left( 1 + \frac{n}{100} \right) = \ln(2). \tag{4}

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

100kln⁡(1+n100)=100ln⁡(2)≃69.3.(5)100k\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:

kln⁡(1+n100)100=100ln⁡(2).(6)k\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)=lim⁡m→∞(1+xm)m.(7)\exp(x) = \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 with the exponential function, thus

kln⁡exp⁡(n)=100ln⁡(2).(8)k\ln\exp(n) = 100\ln(2). \tag{8}

The logarithm is the inverse function of the exponential, therefore

kn=100ln⁡(2).(9)kn = 100\ln(2). \tag{9}

Finally, solving for $k$, we have

k=100ln⁡(2)n≃70n.(10)k = \frac{100\ln(2)}{n} \simeq \frac{70}{n}. \tag{10}

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=ln⁡(2)ln⁡(1+n/100)≃38.85.(11)k = \frac{\ln(2)}{\ln(1+n/100)}\simeq 38.85. \tag{11}
Conclusion:

👍 Very impressive rule of thumb 👍