What is the reason why we use natural logarithm (ln) rather than log to base 10 in specifying functions in econometrics?


8 Answers 8


In the context of linear regression in the social sciences, Gelman and Hill write[1]:

We prefer natural logs (that is, logarithms base $e$) because, as described above, coefficients on the natural-log scale are directly interpretable as approximate proportional differences: with a coefficient of 0.06, a difference of 1 in $x$ corresponds to an approximate 6% difference in $y$, and so forth.

[1] Andrew Gelman and Jennifer Hill (2007). Data Analysis using Regression and Multilevel/Hierarchical Models. Cambridge University Press: Cambridge; New York, pp. 60-61.

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    $\begingroup$ +1: For concrete reason to prefer the natural logarithm. $\endgroup$
    – Neil G
    Aug 6, 2012 at 6:02
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    $\begingroup$ More generally, the exponential function is the only continuous function that is equal to its derivative. $\endgroup$
    – user603
    May 11, 2016 at 14:03
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    $\begingroup$ would this not apply if we apply log10 to the dependent and the independent variable(s)? $\endgroup$
    – cs0815
    Jul 23, 2017 at 18:02
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    $\begingroup$ @cs0815 if you apply the Taylor expansion around the point b $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(b)}{n!}(x-b)^n$$ to the exponential function $f(x)=a^x$, with $f^{(n)}(x) = ln(a)^n a^x$ then you get for the first two terms: $$f(b+x) = f(b) + ln(a)f(b)x + \mathcal{O}(x^2)$$ and the $ln(a)$ term becomes 1 for $a=e$ such that you can use $f(b+x) \sim f(b) (1+x)$, which is however only true for small x. Also you can simply try it out exp(1.06)/exp(1) = 1.0618 and 10^1.06/10^1=1.1418154 $\endgroup$ Oct 17, 2017 at 6:59
  • $\begingroup$ Expanding on @SextusEmpiricus. "small" x are definitely within [-0.5, 0.5] which yields 0.64 (an error of 0.14). See a plot of the relationship between x and the error here: stats.stackexchange.com/a/244237/17459. $\endgroup$ Feb 8 at 9:36

There is no very strong reason for preferring natural logarithms. Suppose we are estimating the model:

ln Y = a + b ln X

The relation between natural (ln) and base 10 (log) logarithms is ln X = 2.303 log X (source). Hence the model is equivalent to:

2.303 log Y = a + 2.303b log X

or, putting a / 2.303 = a*:

log Y = a* + b log X

Either form of the model could be estimated, with equivalent results.

A slight advantage of natural logarithms is that their first differential is simpler: d(ln X)/dX = 1/X, while d(log X)/dX = 1 / ((ln 10)X) (source).

For a source in an econometrics textbook saying that either form of logarithms could be used, see Gujarati, Essentials of Econometrics 3rd edition 2006 p 288.

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    $\begingroup$ The natural log is also useful in a semi-log time series regression since the estimated coefficients can be interpreted as continuously compounded growth rates. $\endgroup$
    – Jason B
    Mar 27, 2012 at 18:20

I think that the natural logarithm is used because the exponential is often used when doing interest/growth calculation.

If you are in continuous time and that you are compounding interests, you will end up having a future value of a certain sum equal to $F(t)=N.e^{rt}$ (where r is the interest rate and N the nominal amount of the sum).

Since you end up with exponential in the calculus, the best way to get rid of it is by using the natural logarithm and if you do the inverse operation, the natural log will give you the time needed to reach a certain growth.

Also, the good thing about logarithms (be it natural or not) is the fact that you can turn multiplications into additions.

As for mathematical explanations of why we end up using an exponential when compounding interest, you can find it here:http://en.wikipedia.org/wiki/Continuously_compounded_interest#Periodic_compounding

Basically, you need to take the limit to have an infinite number of interest rate payment, which ends up being the definition of exponential

Even thought, continuous time is not widely used in real life (you pay your mortgages with monthly payments, not every seconds..), that kind of calculation is often used by quantitative analysts.

  • $\begingroup$ I probably would have given an answer like this. The point that it doesn't matter in modeling is a good one too. We could just as easily use base 2. The difference is only a constant factor $\endgroup$ May 3, 2012 at 18:44
  • $\begingroup$ Surely, you could write it $Nr^t$ (more intuitively), and so are you arguing that the base doesn't matter as Adam Bailey does? $\endgroup$
    – Neil G
    Aug 6, 2012 at 6:16

An additional reason why economists like to use regressions with logarithmic functional forms is an economic one: Coefficients can be understood as elasticities of a Cobb-Douglas function. This function is probably the most common one used among economists to analyze issues regarding microeconomic behaviour (consumers´preferences, technology, production functions) and macroeconomic issues (economic growth). The elasticity term is used to describe the degree of response of a change of a variable with respect to another.


The only reason is that the Taylor expansion, gives an intuitive interpretation of the result.

Let's look at a typical variable used in econometrics a lot, the log difference of GDP: $$\Delta \ln Y_t=\ln Y_t-\ln Y_{t-1}=\ln\frac{Y_{t}}{Y_{t-1}}=\ln\left(1+\frac{\Delta Y_t}{Y_{t-1}}\right)$$ , where $\frac{\Delta Y_t}{Y_{t-1}}$ is GDP growth rate now.

Let's apply Taylor expansion of the log: $$\Delta\ln Y_t\approx\frac{\Delta Y_t}{Y_{t-1}}-\frac 1 2 \left(\frac{\Delta Y_t}{Y_{t-1}}\right)^2+\dots$$ Since, GDP growth rate is usually small, e.g. for USA around 2% lately, we can drop all the higher order terms then we get: $$\Delta\ln Y_t\approx\frac{\Delta Y_t}{Y_{t-1}}$$

So, if you're using the log differences of GDP in the right hand side of the equation, e.g. as an explanatory variable in the regression you may have the following: $$\dots=\dots+\beta\times\Delta\ln Y_t$$ which can be interpreted as "$\beta$ times percentage change in GDP."

Economists like the variables that can be interpreted easily. If you plugged the different log base then the interpretability is weaker. For example, see what happens to the log base 10: $$\dots=\dots+\beta\times\Delta\log_{10} Y_t\\ \approx\dots+\beta\times\frac 1 {\ln(10)}\frac{\Delta Y_t}{Y_{t-1}}$$ This still works, but now you need to divide $\beta$ by some unintuitive number to get the "percentage change" effect interpretation.


Is this unique to economics? The standard normal distribution features an $e^{-{1\over2}x^2}$ in it, and the normal distribution is only one of the large family of exponential distributions that cover a huge swath of statistics. (See GLM's.) It seems like the natural log would be useful in these cases.

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    $\begingroup$ Yes, but the variance term of the normal distribution is equivalent to a change of base of the exponent. Even the way you've drawn it, the base could be $(\sqrt{e})^{-x^2}$ $\endgroup$
    – Neil G
    Aug 6, 2012 at 5:59

There is a good reason to use the log transformation of the variable if you think that the inverse function of logarithm is the exponential function which is a continuous version of conpounding. The economic variable which is growing around 10% at a time can be transformed to the variable with its mean around 10 (plus a constant). You cannot do that with the transformation of logarithm of different base.


Not only in econometrics, using base $e$ is more "natural" in almost every domain, including computer science, where dominated by $0,1$ (where $\log_2$ may be natural).

I would like to use some experiments to show the base $e$ is very natural.

Consider following three functions $f_1(x)=2^x$, $f_{2}(x)=10^x$, $f_3(x)=e^x$, which one seems more natural? Many people may say first two seems better, because $2$ and $10$ are small integers, and $e$ is an irrational number.

However, consider following experiment, we want to investigate the relationship between the function's derivative at $x_0$ and the function value at $x_0$.

We pick two random points, say $1.23$ and $2.34$ and we will use $f_1$ as example.

Here are some facts:

$f_1'(1.23)=1.625894, ~f_1'(2.34)= 3.509423$

$f_1(1.23)=2.34567, ~f_1(2.34)=5.063026$

We see there are some patterns: if we calculate $f_1'(x_0)/f_1(x_0)$, it is a constant: $0.6931472 = \log(2)$.

For another function $f_2$, this constant is $2.302585 = \log(10)$. (I will attach the code for some fun experiments)

So, the natural question to ask is when we can get a simplified results, that do not need to scale with this constant (or this constant is equal to $1.0$)?

The answer is when the base is $e$. Where $f_3'(x)=f_3(x)$. Now, we may think base $e$ is more natural?

some code for fun experiments

# suppose we have some exponential functions 
# because 2 and 10 are small integers f1, f2 seems pretty natural
f1 <- function(x) 2^x
# f2 <- function(x) 10^x

# simple code to calculate numeric gradient
simpleNumDiff <- function(f, x0){

# we try to investigate derv value and function value pattern
# note, they are not equal but have very strong pattern
a1 = simpleNumDiff(f1,1.23)
b1 = f1(1.23)
a2 = simpleNumDiff(f1,2.34)
b2 = f1(2.34)
c(a1/b1,a2/b2) # log(2)
  • $\begingroup$ very nice explain of e in a specific area (computer science)! I mention that in ODE, the equation y'=a y is called one of natural growth rate. $\endgroup$
    – Black Mild
    Jan 30, 2021 at 0:35