Appendix 2B Growth rates and the exponential and logarithm functions

Mary O’Mahony

A2B.1 Introduction

This appendix sets out the principles of estimating the relative change in a variable over time, including estimating annual average growth rates. These underpin many of our key economic indicators, including changes in the levels of GDP, prices and the labour market.

A2B.2 A simple calculation of growth rates

Figure A2B.1 sets out the basic principle of calculating a growth rate. It shows a variable V that changes with time (t). We want to calculate the change in the value of V when we move from t1 to t2. The periods can be years, months, days or even hours, depending on the nature of the variable. We denote changes in discrete time by the Greek symbol Δ (delta). The value of the change in V is calculated as the change in V for the given change in t, given by ΔV = V2V1.

Figure A2B.1 Calculating changes in variables

Calculating changes in variables

Suppose we take a simple example of a variable whose value at a beginning period V1 = 100, and one year later its value is V2 = 120. Then ΔV = V2V1 = 20.

We are usually interested in proportional rates of change. But proportional to which base? Should this be the initial value, the end period value, or possibly an average of the two?

We will return to this example later when we consider using logarithms to calculate growth rates.

A2B.3 Annual average growth rates

annual average growth rate
The annual average growth rate in a variable between year 0 and year T would yield the value of the variable in year T, if starting from the value in year 0, the variable grew at the annual average growth rate each year.

Suppose we now calculate growth over several time periods t0, t1, …, tT and wish to find the annual average growth rate. If time t is measured in years, this is the average growth rate between the starting year t0 and the final year tT.

For a variable Xt, the annual average growth rate between year 0 and year T is given by the following formula:

Annual average growth rate = \(\left(\frac{X_{T}}{X_{0}}\right)^{\frac{1}{T}} − 1\)

How it’s done The formula for the annual average growth rate

Let the starting value of X = X0. Assuming a constant growth rate of g%, the time pattern of X is:

\[X_{1}=X_{0}\left(1 + g\right)\] \[X_{2}=X_{1}\left(1 + g\right)=X_{0}\left(1 + g\right)\left(1 + g\right)=X_{0}\left(1 + g\right)^{2}\] \[X_{3}=X_{2}\left(1 + g\right)=X_{0}\left(1 + g\right)\left(1 + g\right)\left(1 + g\right)=X_{0}\left(1 + g\right)^{3}\]

After three years, we can solve for g as:

\[\left(1 + g\right)^{3}=X_{3}/X_{0}\] \[\left(1 + g\right)=\left(X_{3}/X_{0}\right)^{\left(1/3\right)} \text{ so } g=\left(X_{3}/X_{0}\right)^{\left(1/3\right)} - 1\]

In general, after T years, the formula becomes:

\[g=\left(\frac{X_{T}}{X_{0}}\right)^{\frac{1}{T}} − 1\]

Warning: It may be tempting to just calculate \(\frac{X_{T}}{X_{0}} − 1\) and then divide by the number of periods T, but this would be incorrect as it does not take account of the compounding of growth. Each year the growth rate is calculated on the previous year’s value, which itself is based on a growth from a previous year. For example, suppose expenditure (Xt) by a consumer is initially £500 and grows over the following four years as follows:

\[X_{0}=\text{£}500,X_{1}=\text{£}575,X_{2}=\text{£}810,X_{3}=\text{£}950,X_{4}=\text{£}1,050\]

Using the formula for compound growth derived above, the annual average growth rate is:

\[g=\left(\frac{1,050}{500}\right)^{\frac{1}{4}} − 1 = 0.204 \text{ or 20.4%}\]

Starting with £500 we can use this annual average growth rate to yield the correct end value:

\[\text{£}500(1+0.204)(1+0.204)(1+0.204)(1+0.204)=\text{£}1,050\]

However, suppose instead we calculated the average annual growth rate as the total percentage change between X0 and X4 divided by 4:

\[\lbrack\frac{X_{4}}{X_{0}}−1\rbrack/4=\lbrack\frac{1,050}{500} − 1\rbrack/4=0.275 \text{ or 27.5%.}\]

Using this annual growth rate and, starting with £500, the value after four years is:

\[\text{£}500(1+0.275)(1+0.275)(1+0.275)(1+0.275)=\text{£}1,321\]

This is incorrect because this calculation ignores the compound nature of growth.

Another common error is using the average of each annual growth rate. Figure A2B.2 shows the arithmetic mean of the four annual growth rates which, in this example, is 20.9%. In this case, starting from £500 we get four years later:

\[\text{£}500(1+0.209)(1+0.209)(1+0.209)(1+0.209)=\text{£}1,069\]

which again is the incorrect end period value.

Time (t) Xt Annual growth (%)
0 500  
1 575 0.150
2 810 0.409
3 950 0.173
4 1,050 0.105
  Average 0.209

Figure A2B.2 Annual growth rates for the variable Xt

Annual growth rates for the variable Xt

While using the linear formula \(\frac{X_{t}}{X_{t\space−\space1}}−1\) seems reasonable when calculating annual growth across time, it is not so useful when examining products or ratios of variables. Suppose we need to calculate the growth in labour productivity (LP) defined as output (Y) divided by employment (L). When using the formula above, it turns out that the growth in LP is not equal to the growth in Y minus the growth in L. We will come back to this example when we consider the usefulness of logarithms for calculating growth rates.

Similarly, the same applies when considering the growth in the product of two variables. For example, revenues are equal to prices (P) times quantities sold (Q), but growth in revenues is not the sum of the growth in P and the growth in Q when using the linear formula.

To understand how to tackle these calculations we first need to introduce two functions that are frequently used in economic measurement and analysis: the exponential and logarithmic functions.

A2B.4 Exponential and logarithmic functions

Exponential functions make use of the special number e = 2.71828… which is an irrational number whose digits go on forever without repeating. Mathematically, e is derived as the number that the function \(f(m)=(1+\frac{1}{m})^{m}\) converges to as m gets larger and larger. For example:

If m = 1: \(f(1)=(1+\frac{1}{1})^{1} = 2\)

For m = 2: \(f(2)=(1+\frac{1}{2})^{2}=2.25\)

For m = 10: \(f(10)=(1+\frac{1}{10})^{10}=2.59374\)

For m = 250: \(f(250)=(1+\frac{1}{250})^{250}=2.716924\)

The function gets larger as m increases, but at a decreasing rate, and eventually converges on e = 2.71828. Why should we care about this number? The reason is that the number e has some useful properties when studying growth.

An economic interpretation of e is that it measures the year-end value if £1 is invested at an interest rate of 100% which is continuously compounded. If a person invests £1 at the start of a year at 100% interest, they will have £2 at the end of the year \(((1 + 100\%)^{1} = 2)\).

Suppose instead the interest is compounded every six months (50% twice per year), then they will gain \((1 + 50\%)(1 + 50\%) = (1 + 1/2)^{2} = £2.25\) at the end of the year.

Similarly, if interest is compounded every quarter (25% four times per year), they will gain \((1 + 25\%)(1 + 25\%)(1 + 25\%)(1 + 25\%) = (1 + 1/4)^{4} = £2.44\) at the end of the year.

In the limit, if compounding occurred continuously, they would receive by the end of a year a number that approached e = 2.71828.

An exponential function is where some base number is raised to a power. When the base number is e, this is the natural exponential function, although it is often just called the exponential function.

Examples include:

\[y=e^{t};\space y=e^{5t};\space y=10e^{3t}\text{; or in more general form }y=Ae^{rt}\]

These functions are also frequently denoted by exp():

\[y=\exp(t);\space y=\exp(5t);\space y=10 \exp(3t)\text{; and }y=A \exp(rt)\]

The final example is of particular interest as it is the general expression for interest compounding. Suppose a person invests an amount A for t years at interest rate r, the amount they would receive at the end of the t years is given by \(Ae^{rt}\), provided interest was compounded continuously. In discrete time, this formula would give an approximation.

For example, if £100 is invested at an annual rate of 5% for 10 years, the approximate amount received at the end of the investment period would be:

\[100{e}^{\left(0.05\space×\space10\right)}=\text{£}164.87\]

This compares to the actual return of £162.89 if calculated using the formula \(100 × \left(1 + 0.05\right)^{10}\).

natural logarithm
The power to which e must be raised to attain a particular number. So, if y = ex, then ln(y) = x.

A logarithm is defined as the power to which a base must be raised to attain a particular number. The logarithm to the base e is known as a natural logarithm, often abbreviated to natural log, and is denoted by \(\ln\).

For example, consider the interest rate required for a total return of £164.87 if £100 were invested for 10 years:

\[164.87=100{e}^{\left(r\space×\space10\right)}\]

This simplifies to \(1.6487=e^{10r}\) and therefore \(10r=\ln(1.6487)=0.5\).

In which case, r = 0.05 or 5%.

From this, we can see that the natural logarithm function is the inverse of the exponential function: \(\ln\left(e^{x}\right)=x\). This is an important result. There are some advantages to working with data in logarithms, and this enables data to be transformed into logarithms so analysis can be carried out, and then the results can be converted back into the original units using the exponential function.

As an illustration of the natural logarithm transformation, and why it might be useful in the analysis of growth, consider Figure A2B.3 showing the level of UK real GDP for the post-war period (1948 to 2019). Over this period, the level of real GDP, in constant 2018 prices, increased from £365 billion to £2,173 billion, corresponding to an annual average growth rate of 2.5%. The dashed line on this chart shows a data series that starts with the level of real GDP in 1948 and grows at 2.5% per year to 2019. Although the growth rate of real GDP is constant every year, the actual line itself exhibits the properties of an exponential function.

Now look at Figure A2B.4, which shows the natural log of UK real GDP over the same period (ln(GDP)). The dotted line here shows a linear trend, that starts with the value of ln(GDP) in 1948 and increases by a constant value of 0.025 each year up to 2019. This demonstrates that transforming data using the natural logarithm basically linearises the data. The gradient of the ln(GDP) series between two points provides the respective annual average growth rate, whereas we have seen this would not be the case for the untransformed GDP data.

It is also generally easier to manipulate and model linear data, so the natural log transformation of economic data is frequently used by empirical researchers.

Figure A2B.3 UK real GDP in the post-war period, £ millions

UK real GDP in the post-war period, £ millions

Office for National Statistics – UK National Accounts, The Blue Book: 2020

Figure A2B.4 The natural log of UK real GDP in the post-war period

The natural log of UK real GDP in the post-war period

Office for National Statistics – UK National Accounts, The Blue Book: 2020

A2B.5 Logarithms and growth rates

How it’s done Rules of differentiation

The derivative of a variable y with respect to a variable x describes how much y changes given a very small change in x. It is denoted by dy/dx. The following are some basic rules of differentiation.

\[\begin{alignat*}{3} &y=a+bx && \quad\frac{dy}{dx}=b \\ &y=x^{a} &&\quad\frac{dy}{dx}=ax\quad &&&\text{exponential rule} \\ &y=uv&& \quad\frac{dy}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}\quad &&&\text{product rule} \\ &y=\frac{u}{v} &&\quad\frac{dy}{dx}=\lbrack v\frac{du}{dx} − u\frac{dv}{dx}\rbrack/v^{2}\quad &&&\text{quotient rule} \end{alignat*}\]

When thinking about growth rates. we can consider the differential dy/dt, that is, how the variable y changes over a very small time period (t).

Logarithms are especially useful as they measure proportional growth rates. Using the rules of differentiation, if we differentiate \(y=e^{rt}\) with respect to time t, we get:

\[\frac{dy}{dt}=re^{rt}\]

Given \(y=e^{rt}\), dividing the differential by y gives the proportional growth rate:

\[\frac{1}{y}\frac{dy}{dt}=\frac{1}{e^{rt}}re^{rt}=r\]

The same outcome is arrived at if we were to differentiate ln(y) with respect to time (noting that \(y=e^{rt}\) and \(\ln\,\left(e^{rt}\right)=rt\)).

\[\frac{d\ln\left(y\right)}{dt}=\frac{d(rt)}{t}=r\]

This shows that the derivative of the natural logarithm equals the proportional growth rate in continuous time. This is a very useful property when examining growth. In practice we can use the difference in logs between two adjacent periods as our measure of growth.

To go back to the example at the start of this appendix, with V1 = 100 and V2 = 120, the difference in natural logs is:

\[\ln\left(V_{2}\right) - \ln\left(V_{1}\right) = 4.7875 - 4.6052 = 0.1823 \text{ or 18.23%}\]

This rate is very close to the proportionate growth rate relative to the average of V1 and V2 calculated before:

\[\frac{ΔV}{0.5\left(V_{1}+V_{2}\right)}=\frac{20}{110}=0.1818 \text{ or 18.18%.}\]

Logarithms also have some helpful properties that make them particularly useful functions when considering the growth rate of a product of variables or a ratio of variables. These general properties of logarithms are:

\[\ln\left(uv\right) = \ln\left(u\right) + \ln\left(v\right)\] \[\ln\left(u/v\right) = \ln\left(u\right) - \ln\left(v\right)\] \[\ln\left(u^{a}\right) = a \; \ln\left(u\right)\]

When working in logarithms, you can divide by the number of years to calculate the average annual growth rate, given by \([\ln(X_{T}) - \ln(X_{0})]/T\) or alternatively as \(\ln\left(X_{T}/X_{0}\right)/T\).

To see how, consider the example of UK real GDP growth in the post-war period (1948 to 2019) shown in Figures A2B.3 and A2B.4.

We know that given an annual average growth rate of g:

\[\text{GDP}_{2019} = \text{GDP}_{1948} \left(1 + g\right)^{71}\]

By dividing both sides by GDP1948 and taking the natural log we achieve:

\[\ln\left(\text{GDP}_{2019}/\text{GDP}_{1948}\right) = 71 \, \ln\left(1 + g\right)\]

Therefore, if we use the approximation that for small g, ln(1 + g) ≈ g, then:

\[g ≈ [\ln\left( \text{GDP}_{2019}\right) - \ln\left(\text{GDP}_{1948}\right)]/71\]

Now consider the example of labour productivity LP = (Y/L). There are advantages to using logarithms when dealing with a ratio, as now the growth in labour productivity is equal to the growth in output minus the growth in employment.

To see how, consider the growth in labour productivity from time t − 1 to t using natural logarithms:

\[\text{LP growth}=\ln\left(\frac{Y_{t}}{L_{t}}\right)−\ln\left(\frac{Y_{t\space−\space1}}{L_{t\space−\space1}}\right)\]

Using the rules of logarithms this can be expressed as:

\[=\lbrack \ln\left(Y_{t}\right) − \ln\left(L_{t}\right)\rbrack − \lbrack \ln\left(Y_{t\space−\space1}\right) − \ln\left(L_{t\space−\space1}\right)\rbrack\]

And then rearranged so that:

\[=\lbrack \ln\left(Y_{t}\right) − \ln\left(Y_{t\space−\space1}\right)\rbrack − \lbrack \ln\left(L_{t}\right) − \ln\left(L_{t\space−\space1}\right)\rbrack\]

This shows that when the data is transformed into logarithms, the growth in labour productivity equals output growth minus employment growth.

If growth rates of LP are calculated using the formula \(\frac{LP_{t}}{LP_{t − 1}} − 1\), then it is not equal to output growth minus employment growth.

This can be demonstrated using data for the UK from 1959 to 2019 shown in Figure A2B.5. Figure A2B.5a plots real GDP in constant 2018 prices (GDP), which is the same data in Figure A2B.3, and the total number of jobs (JOBS) as a measure of employment. Figure A2B.5b shows output per job (LP = GDP/JOBS) as a measure of UK labour productivity. We calculate the average annual growth rates using both formulae in Figure A2B.6.

Figure A2B.5 UK labour productivity, 1959 to 2019

UK labour productivity, 1959 to 2019

Office for National Statistics – UK National Accounts, The Blue Book: 2020 and UK Workforce Jobs

Figure A2B.5a Real GDP (£bn) and number of jobs (millions)

Real GDP (£bn) and number of jobs (millions)

Figure A2B.5b Output per job (£)

Output per job (£)

  GDP JOBS LP = (GDP/JOBS) GDP growth minus JOBS growth
[ln(XT) − ln(X0)]/T 2.40 0.55 1.85 1.85
(XT/X0)(1/T) − 1 2.43 0.55 1.86 1.88

Figure A2B.6 Growth in GDP, employment and labour productivity, UK, 1959 to 2019 (% per annum)

Growth in GDP, employment and labour productivity, UK, 1959 to 2019 (% per annum)

For each of these, real GDP (GDP), jobs (JOBS) and labour productivity (LP), the linear and natural log growth rates deviate slightly. In the final column, GDP growth minus JOBS growth is identical to LP growth using the natural log formula, but there is a difference when using the linear formula.

For many calculations the differences between the two formulae are small, but this matters more when variables are growing rapidly.

A2B.6 Summary

When the aim is to measure the proportionate change across a period, it is common to calculate the change relative to the initial period. This is easy to interpret as time moves forward, but care needs to be taken to use the correct formula for average annual growth rates. Finding the total growth and dividing by the number of periods or calculating an average of each period’s growth rate is likely to prove incorrect.

A significant drawback of the linear formula used to calculate average annual growth rates is that it is not additive in products or ratios. An alternative is to use the change in the natural logs of the variable, which is based on proportionate changes in continuous time, and is additive for products and ratios. The simple rules of logarithms can be very useful to researchers in manipulating and modelling economic data.

A final word of caution: since continuous time is not observed, the natural log growth rate is an approximation, using discrete data. It is close to the proportionate growth rate in discrete time if the change is measured relative to the average of the start and end period values.