Interpretation of log transformed predictor

I'm wondering if it makes a difference in interpretation whether only the dependent, both the dependent and independent, or only the independent variables are log transformed.

In the case of

````log(DV) = Intercept + B1*IV + Error 
````

I can interpret the IV as the percent increase but how does this change when I have

````log(DV) = Intercept + B1*log(IV) + Error
````

or when I have

````DV = Intercept + B1*log(IV) + Error?
````
-

I have a feeling the "percent increase" interpretation is not correct but I don't have enough of a grasp to say why exactly.  I hope someone can help....Beyond that, I'd recommend modeling using logs if they help to better establish an X-Y relationship, but reporting selected examples of that relationship using the original variables.  Especially if dealing with an audience that is not too technically savvy. – rolando2 Nov 19 '11 at 15:52
@rolando2: I disagree. If a valid model requires transformation, then a valid interpretation will usually rely on coefficients from the transformed model. It remains the onus of the investigator to appropriately communicate the meaning of those coefficients to the audience. That is, of course, why we get paid such big bucks that out salaries have to be log transformed in the first place. – jthetzel Nov 19 '11 at 19:36
@BigBucks:  Well, look at it this way.  Suppose your audience just can't understand what you mean when you explain that for every change of 1 in the log (base 10) of X, Y will change by b.  But suppose they can understand 3 examples using X values of 10, 100, and 1000.  They at that point will likely catch on to the nonlinear nature of the relationship.  You could still report the overall, log-based b, but giving those examples could make all the difference. – rolando2 Nov 19 '11 at 20:17
....Though now that I read your great explanation below, maybe using those "templates" could help a lot of us clear up these sorts of problems in understanding. – rolando2 Nov 19 '11 at 20:23
Readers here may also want to look at these closely related threads: How to interpret logarithmically transformed coefficients in linear regression, & when-and-why-to-take-the-log-of-a-distribution-of-numbers. – gung Mar 3 at 2:37

2 Answers

Charlie provides a nice, correct explanation. The Statistical Computing site at UCLA has some further examples: http://www.ats.ucla.edu/stat/sas/faq/sas_interpret_log.htm , and http://www.ats.ucla.edu/stat/mult_pkg/faq/general/log_transformed_regression.htm

Just to compliment Charlie's answer, below are specific interpretations of your examples. As always, coefficient interpretations assume that you can defend your model, that the regression diagnostics are satisfactory, and that the data are from a valid study.

Example A: No transformations

````DV = Intercept + B1 * IV + Error 
````

"One unit increase in IV is associated with a (`B1`) unit increase in DV."

Example B: Outcome transformed

````log(DV) = Intercept + B1 * IV + Error 
````

"One unit increase in IV is associated with a (`B1 * 100`) percent increase in DV."

Example C: Exposure transformed

````DV = Intercept + B1 * log(IV) + Error 
````

"One percent increase in IV is associated with a (`B1 / 100`) unit increase in DV."

Example D: Outcome transformed and exposure transformed

````log(DV) = Intercept + B1 * log(IV) + Error 
````

"One percent increase in IV is associated with a (`B1`) percent increase in DV."

-

In the log-log- model, see that $$\begin{equation*}\beta_1 = \frac{\partial \log(y)}{\partial \log(x)}.\end{equation*}$$ Recall that $$\begin{equation*} \frac{\partial \log(y)}{\partial y} = \frac{1}{y} \end{equation*}$$ or $$\begin{equation*} \partial \log(y) = \frac{\partial y}{y}. \end{equation*}$$ Multiplying this latter formulation by 100 gives the percent change in $y$. We have analogous results for $x$.

Using this fact, we can interpret $\beta_1$ as the percent change in $y$ for a 1 percent change in $x$.

Following the same logic, for the level-log model, we have

$$\begin{equation*}\beta_1 = \frac{\partial y}{\partial \log(x)} = 100 \frac{\partial y}{100 \times \partial \log(x)}.\end{equation*}$$ or $\beta_1/100$ is the unit change in $y$ for a one percent change in $x$.

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I never have grasped this. It must be straight forward but I have never seen it... What exactly is \begin{equation*} \partial \log(y) = \frac{\partial y}{y}? \end{equation*} and how do you go from here to a percentage change? – B_Miner Nov 19 '11 at 18:54
All that line does is take the derivative of $\log(y)$ with respect to $y$ and multiply both sides by $\partial y$. We have $\partial y \approx y_1 - y_0$. This fraction, then is the change in $y$ divided by $y$. Multiplied by 100, this is the percent change in $y$. – Charlie Nov 19 '11 at 19:45
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