Transformations in regression

Taking logs (or other transformations) of y or x can straighten curved relationships and stabilise variance, and changes how coefficients are read.

\ln y = b_0 + b_1 x \;\Rightarrow\; \text{each unit of } x \text{ multiplies } y \text{ by } e^{b_1}

Log response: percentage change per unit of x is 100(e^b₁ − 1)%

$b_1$: slope on the log scale; for small |b₁|, 100·b₁ % is a good approximation
$e^{b_1}$: multiplicative effect on (the median of) y

y = b_0 + b_1 \ln x

Log predictor: a 1% increase in x changes y by about b₁/100

$b_1$: change in y when x is multiplied by e (≈ 2.718)

\ln y = b_0 + b_1 \ln x

Log–log: b₁ is an elasticity (≈ % change in y per 1% change in x)

$b_1$: elasticity

When to use

Right-skewed positive responses (prices, incomes, counts of sales), growth processes, residual funnels that widen with the fitted value, or curved scatter plots.
Logs need strictly positive values.

How to interpret

Log y: effects are multiplicative (percentages). Back-transforming a prediction, $e^{\hat y}$ , estimates the median of y (not the mean) when log-errors are symmetric.
Choose the transformation from the residual plot and subject knowledge, then re-check the residuals.

Common mistakes

Reading a log-scale slope as a change in units of y.
Comparing $R^2$ between a model for y and a model for ln y — they explain different quantities.
Using 100·b as the percentage when b is large (b = 0.5 means +65%, not +50%).
Adding an arbitrary constant before logging zeros without saying so.

Example

y = 10, 12, 15, 18, 22, 27 at x = 0…5. A straight line leaves U-shaped residuals; regressing ln y on x gives $\ln\hat y = 2.298 + 0.1991x$ ( $R^2 = 0.9995$ , residuals patternless).

Each step in x multiplies y by $e^{0.1991} = 1.220$ : a 22.0% increase (the 100·b shortcut says 19.9%).
Predicted y at x = 0: $e^{2.298} = 9.95$ .