In log-log regression why are the $\beta$1 coefficients the same for non natural logs but the $\beta$0 coefficients different? Can the $\beta$1 coefficients for non natural log-log regression still be interpreted as a 'One percent increase in IV is associated with a $\beta$1 percent increase in DV' Here, it would be for a 1% change in body mass we'd expect to see a 0.6518 % change in brain mass.

body = c(62000,277,5000000,160000,28000,960,200,60000000,37000000,2500)
brain = c(1400,7.5,6000,1700,46.2,7.4,3,6000,7820,12)

fit1 = lm(log(brain)~log(body))

fit2 = lm(log10(brain)~log10(body))

fit3 = lm(log2(brain)~log2(body))

EDIT: Is the maths for the $log_{10}$ case the following (based on Interpretation of log transformed predictor):

In the log-log- model, see that $$\begin{equation*}\beta_1 = \frac{\partial \log_{10}(y)}{\partial \log_{10}(x)}.\end{equation*}$$ which gives $$\begin{equation*} \frac{\partial \log_{10}(y)}{\partial y} = \frac{1}{y} \end{equation*}$$ or $$\begin{equation*} \partial \log_{10}(y) = \frac{\partial y}{y}. \end{equation*}$$


2 Answers 2


If you understand how to convert units--such as kilograms to pounds or meters to feet--then you will understand perfectly what is going on here, too, because it involves a simple change of units. (For more about this, please see Gung's answer to How will changing the units of explanatory variables affect a regression model?)

The "base" of a logarithm is its units of measurement. That is, changing bases amounts to multiplication by a constant:

$$\log_b(x) = c\, \log(x)\tag{1}$$

where $c = 1/(\log b)$.

This gives you all you need to answer the questions. Begin with the first model,

$$\log \text{brain} = \alpha + \beta\,\log\text{body}.$$

Both sides use natural logs: that's their common unit of measurement for logarithms. Specifically, $\alpha$ is measured in "nats" and $\beta$ is "nats per nat": it is unitless. Compare this to the second model using $(1)$

$$c \log \text{brain} = \log_{10}\text{brain} = \alpha_{10} + \beta_{10}\,\log_{10}\text{body} = \alpha_{10} + \beta_{10}\,c \log\text{body}.$$

Both sides use common logs. Changing back to the original units by dividing through by $c$ yields

$$\log \text{brain} = \frac{1}{c}\alpha_{10} + \beta_{10}\,\log\text{body}.$$

Comparing this to the first model shows immediately that

$$\alpha = \frac{1}{c}\alpha_{10},\quad \beta = \beta_{10}.$$

In other words,

the slopes do not change (they cannot, since they are unitless) but the intercept must have its units converted (from nats to a common log) to match those of the new base of logarithms. The same relationship must hold among the estimated coefficients, too.

In particular,

the invariance of the slopes shows you do not need to change your interpretation of them.

As an example, consider the estimates in your code. It reports $\hat\alpha = -1.8980$ and $\hat\beta = 0.6518$. We therefore anticipate that $\hat\alpha_{10} = (1/c)(-1.8980)$ where $c=\log{10} = 2.30\ldots$. The value works out to $\hat \alpha_{10} = (1/2.30\ldots)(-1.8980) = -0.8243$. Sure enough, that's precisely what the second model outputs for the intercept (and the two slope estimates are the same).


The key issue here is that you have the same base of the log on both sides. So an estimated $\beta_1=1$ tells you that if you multiple body by the base of the logs you multiply brain by $\beta_1 \times$ the base of the logs. Since the base of the logs is the same on both sides it factors out.

If you raise the base of the logs to the power of the intercept you will find all three are the same.

  • $\begingroup$ Thanks for your answer, so does the maths follow what I've set out in my edit for the log10 case? $\endgroup$
    – adkane
    Commented Jan 9, 2017 at 17:31
  • $\begingroup$ I think the replies given there were sound, yes. $\endgroup$
    – mdewey
    Commented Jan 9, 2017 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.