I have read many threads here on how to interpret coefficients in a regression where the predictor and the dependent variable are log-transformed. Most give an answer for a one or ten percent change. However, I am not sure whether I am right about interpreting other changes.

I have a beta coefficient of $-0.5057$. Given the standard interpretation (as elasticity), a one percent increase in the predictor variable ($X$) leads to an decrease of $0.5057$ percent in the dependent variable ($Y$). Alternatively a ten percent increase in $X$ leads to a $5.057$ percent decrease in $Y$.

I found the formulas $1.1^{-0.5057}$ and $\exp(-0.5057(\log(1.1)))$ to calculate the affect of any change in $X$ on $Y$. Are these formulas right because I get for a ten percent increase slightly different results than the $5.057$ decrease.

Furthermore, is it right that the percentage values for the changes in $Y$ do not increase proportionally with increases in $X$? If a one percent increase in $X$ leads to an decrease of $0.5057$ percent in $Y$, it is wrong to assume that a $50$ percent increase in $X$ leads to a $(50 \times 0.5057)$ decrease in $Y$?

$\exp(-0.5057(\log(1.5)))= 0.8146$. Given the difference between $1$ and this value, it suggests that $Y$ decreases by $18.5$ percent. Am I right?

The formulas mentioned above suggest so. Is that because of the nonlinearity in the model?

  • 1
    $\begingroup$ If your discomfort is with the wordiness of interpretting findings along the lines of "we observed an $e^\beta$-fold difference in $y$ for a $e$ fold difference in $\beta$" (or whatever the appropriate base is, 2 is often a better choice), I don't see any problem with directly interpretting $\log(y)$ and $\log(x)$ directly as measures of interest, for instance "We observed a $\beta$ difference in the log of y for a 1-log difference in x. $\endgroup$
    – AdamO
    Commented Jul 6, 2016 at 21:43

1 Answer 1


The question concerns models of the form

$$\log(y) = \cdots + \beta \log(x) + \cdots$$

(where none of the omitted terms involves anything that changes with $x$). When we change $x$ by $100\delta\%$ we multiply it by $1+\delta$. According to the laws of logarithms, when $1 + \delta \gt 0$,

$$\log(x(1+\delta)) = \log(x) + \log(1 + \delta).$$

Therefore if such a change in $x$ changes $y$ to $y^\prime$,

$$\eqalign{ \log(y^\prime) &= \cdots + \beta \log(x(1+\delta)) + \cdots \\ &= \cdots + \beta\left(\log(x) + \log(1 + \delta)\right) + \cdots \\ &= \cdots + \beta\log(x) + \beta\log(1 + \delta) + \cdots . }$$

The change in $\log(y)$ is $$\log(y^\prime) - \log(y) = \beta\log(1 + \delta).$$

When $\delta$ is nearly zero (say, $10\%$ or smaller in size), $\log(1 + \delta) \approx \delta$ is a good approximation. When in turn $\beta\delta$ is also close to zero, this is the basis for the approximate interpretation "a $\delta$ percent change in $x$ corresponds to a $\beta\delta$ percent change in $y$."

For larger $\delta$ or $\beta$, however, this approximation fails. The fully general relationship is obtained by noting that adding $\beta\log(1+\delta)$ to $\log{y}$ is tantamount to multiplying $y$ by

$$\exp(\beta\log(1+\delta)) = (1+\delta)^\beta.$$

Therefore, when working with logarithms, think multiplicatively. We may memorialize the result of this analysis with a simple rule:

When $x$ is multiplied by any positive amount $c$, $y$ is multiplied by $c^\beta$.

In other words, log-log relationship are power relationships. Let's look at some examples:

  • When $\beta=2$, multiplying $x$ by $c$ multiplies $y$ by $c^2$. For instance, tripling $x$ will multiply $y$ by $9$.

  • When $\beta=1/3$, multiplying $x$ by $c$ multiplies $y$ by $c^{1/3} = \sqrt[3]{c}$. For instance, doubling $x$ will only multiply $y$ by $\sqrt[3]{2}\approx 1.26$.

  • When $\beta = -1$, multiplying $x$ by $c$ multiplies $y$ by $c^{-1} = 1/c$; that is, $y$ is divided by $c$.

  • $\begingroup$ Are you exact with the numbers? For example ln(9) = 2.2, not 2.0. So shouldn't it be that multiplying x by 2.2 we triple y. Or multiplying x by 2.0 we multiply y by 7.4? $\endgroup$ Commented Feb 6, 2020 at 21:46
  • $\begingroup$ @PrzemyslawRemin You seem to be confusing power relationships with exponential relationships. My examples are exact except where I indicate approximations with $\approx.$ $\endgroup$
    – whuber
    Commented Feb 6, 2020 at 23:18
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    $\begingroup$ I am happy to learn that I was wrong:-) I like you simple rule :-) $\endgroup$ Commented Feb 6, 2020 at 23:22

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