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Playing with logs

In applied work, we often take logs to make relationships closer to linear, reduce the influence of extreme values, and make coefficients easier to compare across contexts. But once a variable is in logs, the meaning of β1\beta_1β1​ changes: the same estimated number can represent a unit change, a percent change, or an elasticity depending on whether YYY and XXX are in levels or logs. This note is a quick reference for translating regression output into plain language.

For example, suppose two schools use different versions of an exam. One is scored from 0 to 100, and another is scored from 200 to 800. A 10-point increase does not mean the same thing across the two scales. Reporting changes in percent terms (or, more generally, in relative terms) makes comparisons more interpretable across contexts where units and scales differ.

1. Level–Level model

Model: Y=β0+β1X+uY = \beta_0 + \beta_1 X + uY=β0​+β1​X+u

Interpretation of β1\beta_1β1​:

A one-unit increase in XXX is associated with a β1\beta_1β1​ increase in units of YYY.

Example:

If YYY is income (in dollars) and XXX is years of education, and β1=2000\beta_1 = 2000β1​=2000, then one more year of education is associated with $2,000 higher income.

2. Log–Level model (log YYY, level XXX)

Model: ln⁡(Y)=β0+β1X+u\ln(Y) = \beta_0 + \beta_1 X + uln(Y)=β0​+β1​X+u

Interpretation of β1\beta_1β1​ (approximate):

A one-unit increase in XXX is associated with approximately a (100×β1)%(100 \times \beta_1)\%(100×β1​)% change in YYY

More precise interpretation:

Percent change in YYY is 100×(eβ1−1)100\times (e^{\beta_1} - 1)100×(eβ1​−1)

Example:

If β1=0.05\beta_1 = 0.05β1​=0.05, then a one-unit increase in XXX increases YYY by approximately 5%

3. Level–Log model (level YYY, log XXX)

Model: Y=β0+β1ln⁡(X)+uY = \beta_0 + \beta_1 \ln(X) + uY=β0​+β1​ln(X)+u

Interpretation of β1\beta_1β1​:

A 1%1\%1% increase in XXX is associated with a β1100\frac{\beta_1}{100}100β1​​ unit increase in Y.

Example:

If β1\beta_1β1​ = 40, then a 1%1\%1% increase in XXX increases YYY by 0.40.40.4 units of Y

4. Log–Log model (log YYY, log XXX)

Model: ln⁡(Y)=β0+β1ln⁡(X)+u\ln(Y) = \beta_0 + \beta_1 \ln(X) + uln(Y)=β0​+β1​ln(X)+u

Interpretation of β1\beta_1β1​:

β1\beta_1β1​ is an elasticity. A 1%1\%1% increase in XXX is associated with a β1%\beta_1\%β1​% change in Y

Example:

If β1=0.8\beta_1 = 0.8β1​=0.8, then a 1%1\%1% increase in XXX increases YYY by 0.8%.

Practice exercises

  1. A researcher is studying how an additional year of schooling is associated with earnings. Wages are often right-skewed and measured in dollars, so using ln⁡(wage)\ln(\\\text{wage})ln(wage) helps reduce the influence of very high earners and makes the coefficient easier to interpret in percentage terms. The researcher estimates:
    1. ln⁡(wage)=1.2+0.07 education\ln(\text{wage}) = 1.2 + 0.07\,\text{education}ln(wage)=1.2+0.07education
    2. Interpret the coefficient on education.
    3. What is the approximate percent increase in wage from one additional year of education?
    4. What is the exact percent increase?
    5. ‣
      Answer
  2. A researcher is studying how study time is associated with test performance. Study time often has diminishing returns, so using ln⁡(study_hours)\ln(\text{study\_hours})ln(study_hours) can capture a relationship where percent increases in studying have similar associations with scores. They estimate
    1. test_score=500+30 ln⁡(study_hours)\text{test\_score} = 500 + 30\,\ln(\text{study\_hours})test_score=500+30ln(study_hours)
    2. Interpret the coefficient on ln(study hours)ln(study\ hours)ln(study hours).
    3. What happens to test scores if study hours increase by 10%?
    4. ‣
      Answer
  1. A researcher wants to estimate the responsiveness of sales to advertising. Using logs makes the coefficient an elasticity. They estimate:
    1. ln⁡(sales)=2.5+1.3 ln⁡(advertising)\ln(\text{sales}) = 2.5 + 1.3\,\ln(\text{advertising})ln(sales)=2.5+1.3ln(advertising)
    2. Interpret the coefficient 1.3.
    3. If advertising increases by 5%, what is the predicted percent change in sales?
    4. ‣
      Answer
  1. A researcher is studying returns to labor market experience. A levels model answers the question in dollars, while a log-income model answers it in percentage terms. Suppose you estimate two models:
    • Model A: income=10000+500 experience\text{income} = 10000 + 500\,\text{experience}income=10000+500experience
    • Model B: ln⁡(income)=8+0.04 experience\ln(\text{income}) = 8 + 0.04\,\text{experience}ln(income)=8+0.04experience
    • Interpret the coefficient on experience in both models.
    • Which model implies diminishing returns to experience? Why?
    • If experience increases by 5 years, how much does income increase in each model?
    • ‣
      Answer
  1. A researcher wants to understand how housing characteristics relate to sale prices. They use logs to interpret effects in percentage terms and compare across markets with different price levels and typical home sizes. A researcher estimates:
    1. ln⁡(house_price)=10+0.6 ln⁡(square_feet)+0.03 bedrooms\ln(\text{house\_price}) = 10 + 0.6\,\ln(\text{square\_feet}) + 0.03\,\text{bedrooms}ln(house_price)=10+0.6ln(square_feet)+0.03bedrooms
    2. Interpret both coefficients carefully.
    3. What is the change in square feet such that the change in price is the same as adding two bedrooms?
    4. ‣
      Answer