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By Daniil Tarakanov Updated Oct 16, 20262026 curriculum

Functional forms: log-lin, lin-log and log-log models

Take logs to straighten a curve. Log-lin (ln Y on X) fits steady percentage growth: the slope is the relative change in Y per unit of X. Lin-log (Y on ln X) fits fading effects: the slope is the change in Y for a relative change in X. Log-log (ln Y on ln X) gives an elasticity.

6 min cram, 34 min full read. Based on the 2026 CFA® Program Level I curriculum, Quantitative Methods, Learning Module 10. This is the 2026 version of the lesson. Sitting in February 2027 or later? Read the updated 2027 version.

Tailor this to you: take the self-assessment

Before this: The four regression assumptions and what residual plots reveal

On the exam

Describe different functional forms of simple linear regressions

Describe the log-lin, lin-log and log-log functional forms, interpret their slopes and pick the right form. Expect to convert a log-model prediction back with e^x, or to read a log-log slope as an elasticity. Traps: reading a log-lin slope as an absolute change, forgetting to exponentiate.

How deeply to learn each part

Know the idea: Understand it in general. It frames other topics, but you will not calculate with it.
Use it: Understand it well enough to solve a question with it. Recognize when it applies.
Memorize: Learn it by heart. The exam will not give it to you.

You know this one. The full lesson is folded away. The takeaways, formulas and checkpoint below are your quick refresher, and the checkpoint is a fair test of whether you still do.

You tested strong here. The full lesson is folded away. The takeaways, formulas and checkpoint below are your quick refresher, and the checkpoint is a fair test of whether you still know it.

What if the relationship is not a straight line?

A straight line fits some relationships badly. If a company’s revenue compounds at a steady percentage, a scatter against time bends upward, and the residuals of a linear fit trace a U. The remedy is to change the functional form : replace Y, X or both with their logarithms. The natural logarithm transformation turns multiplicative growth into additive steps, which a straight line handles.

The regression machinery is unchanged. Everything from the earlier lessons (least squares, R², t-tests, the standard error of estimate or SEE) works on the transformed variables. What changes is what the slope means and how you predict.

Choosing a functional form

Which side of the equation is in logs?logs straighten a curved scatter

Neither: lin-lin Use it the ordinary straight-line model
Y only: log-lin Memorize ln Y = b₀ + b₁X
- Use for constant percentage growth over time
- Slope relative change in Y per unit X
X only: lin-log Memorize Y = b₀ + b₁ ln X
- Use for diminishing effects
- Slope change in Y per relative change in X
Both: log-log Memorize ln Y = b₀ + b₁ ln X
- Use for constant-elasticity relations
- Slope elasticity, % change in Y per 1% in X

What is the log-lin model?

In the log-lin model only Y is in logs.

Log-lin model Memorize

\ln Y_i = b_0 + b_1 X_i + \varepsilon_i

The slope is the relative change in Y for a one-unit change in X. For small values, $b_1 \times 100$ is the percentage growth; the exact growth is $e^{b_1} - 1$ . This is the model for compound growth over time, with X as the time period.

To predict Y you must reverse the log:

Converting a log-lin prediction back to Y Memorize

Y = e^{\,b_0 + b_1 X}

Worked example Revenue that compounds

A fit of ln(Revenue) on the year number $t$ gives $\ln(\text{Revenue}) = 4.2 + 0.065t$ , with revenue in millions.

Growth: $e^{0.065} - 1 = 6.72\%$ a year (roughly the 6.5% that the slope suggests).
Prediction at $t = 10$ : $\ln \text{Revenue} = 4.2 + 0.065 \times 10 = 4.85$ .
Revenue $= e^{4.85} =$ 127.7 million.

Step 2 and 3 on your calculator

0.065×10+4.2=2NDe^x 127.74

2ND second function (the shifted label above a key) · e^x e to the power x

0.065ENTER10×4.2+ge^x 127.74

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · e^x e to the power x

Rearrange itLog-lin prediction: solve for any one

The question asks for

A log-lin model is ln Y = 4.2 + 0.065X. For X = 10, the predicted Y is closest to:

Intercept, b₀Slope, b₁Value of X (here, the year number)

Y = e^{b_{0} + b_{1} X}

e^(4.2 + 0.065 × 10) = e^4.85 = 127.7404

Predicted Y127.7404

On your calculator

0.065×10+4.2=2NDe^x

2ND second function (the shifted label above a key) · e^x e to the power x

0.065ENTER10×4.2+ge^x

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · e^x e to the power x

What is the lin-log model?

In the lin-log model only X is in logs. Equal percentage rises in X give equal absolute rises in Y, so the effect of each extra unit of X fades as X grows.

Lin-log model Memorize

Y_i = b_0 + b_1 \ln X_i + \varepsilon_i

The slope is the change in Y for a relative change in X: a 1% rise in X changes Y by about $b_1 \times 0.01$ .

What is the log-log model?

In the log-log model both variables are in logs.

Log-log model Memorize

\ln Y_i = b_0 + b_1 \ln X_i + \varepsilon_i

The slope is an elasticity : the percentage change in Y for a 1% change in X. It is the standard way to estimate a demand curve with constant elasticity.

Rearrange itLog-log elasticity: solve for any one

The question asks for

In a log-log model the slope is 0.6. A 10% rise in X is associated with a change in Y closest to:

Elasticity (log-log slope), b₁Change in X%

%Δ Y \approx b_{1} \times %Δ X

0.6 × 10% = 6%

Approximate change in Y6.00%

On your calculator

0.6×10=

0.6ENTER10×

ENTER enter (on the HP, puts the number on the stack)

Loading the interactive

How do you choose the form?

Look at the scatter and the residual plot of a linear fit. A curved pattern in the residuals is the signal. Try the transformation that matches the economics: compounding growth suggests log-lin, fading effects suggest lin-log, constant percentage responses suggest log-log. The right form leaves residuals that look like random noise.

You can compare candidate models using R², F and SEE, but only when the dependent variable is the same. The R² of a model for ln Y and the R² of a model for Y describe the variation of different things, so they cannot be compared directly.

Exam tip Name the form by where the logs are: log on Y only is log-lin, on X only is lin-log, on both is log-log. Then say what the slope is a change in.

Go deeper Why compound growth is linear in logs

If revenue grows by the same factor each year, $Y_t = Y_0 (1+g)^t$ , then $\ln Y_t = \ln Y_0 + t\ln(1+g)$ . That is a straight line in $t$ with slope $\ln(1+g)$ , the continuously compounded growth rate you met in the rates and returns lessons. Regressing the log of revenue on time estimates that rate directly, and $e^{b_1} - 1$ converts it to the ordinary annual growth.

Go deeper Elasticities in economics

The price elasticity of demand is the percentage change in quantity for a 1% change in price. A log-log demand regression estimates it as a constant across all prices, which is why the form appears in economics. A slope of −1.2 means a 1% price rise cuts quantity by about 1.2%, and demand is elastic.

Memorize this

MemorizeThree functional forms

Model	Equation	Slope means	Fits
Log-lin	$\ln Y = b_0 + b_1X$	relative change in Y per unit X ( $\approx b_1 \times 100\%$ )	steady percentage growth
Lin-log	$Y = b_0 + b_1\ln X$	change in Y for a relative change in X (1% gives $b_1 \times 0.01$ )	effects that fade as X grows
Log-log	$\ln Y = b_0 + b_1\ln X$	elasticity: % change in Y per 1% change in X	constant-elasticity relations

Memory hook Read the model name as "Y-side, X-side": the log sits on the side that is measured in percent. Log on Y means Y moves in percent; log on X means X moves in percent. Log on both: percent per percent.

Know it cold A prediction from a model with ln Y must be exponentiated: Y = e^(predicted ln Y). Compare R² only between models with the same dependent variable.

Choosing a functional form

Which side of the equation is in logs?logs straighten a curved scatter

Neither: lin-lin Use it the ordinary straight-line model
Y only: log-lin Memorize ln Y = b₀ + b₁X
- Use for constant percentage growth over time
- Slope relative change in Y per unit X
X only: lin-log Memorize Y = b₀ + b₁ ln X
- Use for diminishing effects
- Slope change in Y per relative change in X
Both: log-log Memorize ln Y = b₀ + b₁ ln X
- Use for constant-elasticity relations
- Slope elasticity, % change in Y per 1% in X

Log-lin model

\ln Y_i = b_0 + b_1 X_i + \varepsilon_i

Quick example

ln Y = 1.0 + 0.10X. At X = 3, ln Y = 1.30, so Y = e^1.30 = 3.67. The slope says Y grows about 10% per unit of X.

On your calculator

0.1×3+1=2NDe^x 3.6693

2ND second function (the shifted label above a key) · e^x e to the power x

0.1ENTER3×1+ge^x 3.6693

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · e^x e to the power x

Lin-log model

Y_i = b_0 + b_1 \ln X_i + \varepsilon_i

Quick example

Y = 4 + 2 ln X. X rises from 50 to 60. Y changes by 2 × ln(60 ÷ 50) = 2 × 0.1823 = 0.36.

On your calculator

60÷50=LN×2= 0.3646

LN natural logarithm

60ENTER50÷gLN2× 0.3646

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · LN natural logarithm

Log-log model

\ln Y_i = b_0 + b_1 \ln X_i + \varepsilon_i

Quick example

Price elasticity of demand b₁ = −1.2. A 5% price rise goes with a change in quantity of about −1.2 × 5% = −6%.

On your calculator

1.2+|-×5= −6

+|- change sign

1.2CHSENTER5× −6

CHS change sign · ENTER enter (on the HP, puts the number on the stack)

Converting a log-lin prediction back to Y

Y = e^{\,b_0 + b_1 X}

Quick example

ln Y = 2.0 + 0.05X. At X = 20: ln Y = 3.0, so Y = e^3.0 = 20.09.

On your calculator

0.05×20+2=2NDe^x 20.0855

2ND second function (the shifted label above a key) · e^x e to the power x

0.05ENTER20×2+ge^x 20.0855

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · e^x e to the power x

How it is asked

Predict from a log-lin model

ln Y = 2.1 + 0.08X. Predicted Y at X = 5?

ln Y = 2.5, so Y = e^2.5 = 12.18. Trap: 2.50 is the log, not Y.
Elasticity

ln Sales = 3.0 + 0.6 ln Price index. Price index up 10%?

Sales up about 0.6 × 10% = 6%.
Which model fits growth?

Revenue grows at a roughly constant percentage a year. Best model?

Log-lin: ln(revenue) on time.
Lin-log change

Y = 4.0 + 2.5 ln X. X rises from 80 to 100. Change in Y?

2.5 × ln(100 ÷ 80) = 0.56.

Exam traps

Reading a log-lin slope as an absolute change in Y: it is a relative (percentage) change.
Forgetting to exponentiate a prediction from a model with ln Y.
Comparing R² across models with different dependent variables (Y versus ln Y).
Reading a log-log slope as a change in units: it is a percent-per-percent elasticity.

Checkpoint

4 exam-style questions. About 90 seconds each.

1A log-lin model is ln Y = 2.1 + 0.08X. For X = 5, the predicted Y is closest to:
2In the model ln(Sales) = 3.0 + 0.6 ln(Price index), a 10% rise in the price index is associated with sales closest to:
3Revenue that grows at a roughly constant percentage each year is best modeled with a:
4A lin-log model is Y = 4.0 + 2.5 ln X. When X rises from 80 to 100, the predicted change in Y is closest to:

Key takeaways

Log-lin: relative change in Y per unit of X (growth). Lin-log: change in Y per relative change in X. Log-log: elasticity.
Exponentiate predictions from models whose dependent variable is ln Y.
Pick the form that leaves boring residuals, and compare fit only with the same dependent variable.

Log-lin model

\ln Y_i = b_0 + b_1 X_i + \varepsilon_i

Quick example

ln Y = 1.0 + 0.10X. At X = 3, ln Y = 1.30, so Y = e^1.30 = 3.67. The slope says Y grows about 10% per unit of X.

On your calculator

0.1×3+1=2NDe^x 3.6693

2ND second function (the shifted label above a key) · e^x e to the power x

0.1ENTER3×1+ge^x 3.6693

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · e^x e to the power x

Lin-log model

Y_i = b_0 + b_1 \ln X_i + \varepsilon_i

Quick example

Y = 4 + 2 ln X. X rises from 50 to 60. Y changes by 2 × ln(60 ÷ 50) = 2 × 0.1823 = 0.36.

On your calculator

60÷50=LN×2= 0.3646

LN natural logarithm

60ENTER50÷gLN2× 0.3646

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · LN natural logarithm

Log-log model

\ln Y_i = b_0 + b_1 \ln X_i + \varepsilon_i

Quick example

Price elasticity of demand b₁ = −1.2. A 5% price rise goes with a change in quantity of about −1.2 × 5% = −6%.

On your calculator

1.2+|-×5= −6

+|- change sign

1.2CHSENTER5× −6

CHS change sign · ENTER enter (on the HP, puts the number on the stack)

Converting a log-lin prediction back to Y

Y = e^{\,b_0 + b_1 X}

Quick example

ln Y = 2.0 + 0.05X. At X = 20: ln Y = 3.0, so Y = e^3.0 = 20.09.

On your calculator

0.05×20+2=2NDe^x 20.0855

2ND second function (the shifted label above a key) · e^x e to the power x

0.05ENTER20×2+ge^x 20.0855

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · e^x e to the power x

Cheat sheet · PDF · 4 A4 pages

Simple Linear Regression on 4 pages

What to memorize, the concept maps, every formula with a worked example and keystrokes for the TI BA II Plus and HP 12C, how each idea is asked, and the traps. Print it, or keep it on your phone.

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Where this comes back

Simple Linear Regression Quantitative Methods
Introduction to Financial Statement Modeling Financial Statement Analysis

Questions candidates ask

How do you interpret a log-lin regression coefficient?

In ln Y = b₀ + b₁X, the slope is the relative change in Y for a one-unit rise in X. A slope of 0.065 means Y grows about 6.5% per unit of X (exactly e^0.065 − 1 = 6.7%). It is not an absolute change in Y.

What does the slope mean in a log-log model?

It is an elasticity: the approximate percentage change in Y for a 1% change in X. With ln Sales = 1.8 + 0.4 ln Advertising, a 1% rise in advertising goes with about a 0.4% rise in sales.

When should you use a log transformation in regression?

When a scatter or residual plot is curved or fanning, and the relation is better described in percentages: compounding growth (log-lin), effects that fade (lin-log) or constant elasticities (log-log). Choose the form that leaves patternless residuals.

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