Math Core

Many real relationships are curved: populations grow exponentially, and the period of a pendulum grows with the square root of its length. The least-squares line can't fit a curve directly. But if you transform one or both variables, for example by taking logarithms, a curved pattern can become a straight one. Then all the linear tools from this unit work again, and you undo the transformation at the end to make predictions.

When a line isn't enough

A lab technician counts the bacteria in a culture (in thousands) every hour.

Hour, $x$	$0$	$1$	$2$	$3$	$4$	$5$
Count, $y$	$20$	$31$	$44$	$68$	$101$	$150$

x-axis: hour. y-axis: bacteria count (thousands). The LSRL ŷ = 5.857 + 25.257x misses the curve.Open in grapher →

A linear fit gives $\hat{y} = 5.857 + 25.257x$ with $r^2 \approx 0.926$ . That $r^2$ looks respectable, but the residuals are $14.1, -0.1, -12.4, -13.6, -5.9, 17.9$ : positive, then negative, then positive. That U-shaped residual pattern says the linear model is wrong. The counts don't go up by the same amount each hour; they go up by roughly the same factor (about $1.5$ times). That is exponential growth.

Linearizing exponential growth: log of y

If $y = a \cdot b^x$ is exponential, then taking the base- $10$ logarithm of both sides gives

\log y = \log a + x \log b.

That's a linear equation in $x$ with intercept $\log a$ and slope $\log b$ . So if the data are roughly exponential, a scatterplot of $\log y$ against $x$ should be roughly linear.

Hour, $x$	$0$	$1$	$2$	$3$	$4$	$5$
$\log y$	$1.301$	$1.491$	$1.643$	$1.833$	$2.004$	$2.176$

x-axis: hour. y-axis: log(count). The transformed points lie almost exactly on a line.Open in grapher →

The LSRL for the transformed data is

\widehat{\log y} = 1.306 + 0.174x, \qquad r^2 \approx 0.9995.

The residual plot for the transformed data shows no pattern:

Residuals from the log model. x-axis: hour. y-axis: residual in log(count). Random scatter.Open in grapher →

Choosing a transformation

Exponential pattern ( $y$ multiplies by a constant factor as $x$ goes up by a constant amount): plot $\log y$ against $x$ .
Power pattern ( $y = a x^p$ ): plot $\log y$ against $\log x$ .
Other curves: try $\sqrt{y}$ , $y^2$ , $\dfrac{1}{x}$ or similar re-expressions.

Whichever you try, judge success the same way as always: the transformed scatterplot should look linear and the residual plot should show no leftover pattern.

Making predictions: undo the transformation

The model predicts $\log y$ , not $y$ , so the last step is always to reverse the transformation.

Worked example: Predicting from an exponential model

Use $\widehat{\log y} = 1.306 + 0.174x$ to predict the bacteria count at $x = 3.5$ hours.

First predict the logarithm:

\widehat{\log y} = 1.306 + 0.174(3.5) = 1.306 + 0.609 = 1.915.

Then undo the log by raising $10$ to that power:

\hat{y} = 10^{1.915} \approx 82.2.

The predicted count is about $82{,}200$ bacteria ( $82.2$ thousand). That sits sensibly between the observed $68$ at hour $3$ and $101$ at hour $4$ .

The slope also has a multiplicative meaning. Each additional hour adds $0.174$ to the predicted $\log y$ , which multiplies the predicted count by $10^{0.174} \approx 1.49$ . So the model says the culture grows by about $49\%$ per hour.

Common mistake

Don't stop at $\widehat{\log y}$ . A prediction of " $1.915$ bacteria" is the most common error on transformation questions. Always ask: is my answer in the original units? If the model used $\ln y$ instead of $\log y$ , undo it with $e^{(\cdot)}$ , not $10^{(\cdot)}$ .

Linearizing a power model: log of both

A physics class times the period $T$ (seconds) of a pendulum for several lengths $L$ (centimeters).

$L$ (cm)	$10$	$20$	$40$	$60$	$80$	$100$
$T$ (s)	$0.63$	$0.90$	$1.27$	$1.55$	$1.79$	$2.01$

x-axis: length (cm). y-axis: period (s). The pattern bends and levels off.Open in grapher →

The plot rises but bends downward, the shape of a power function with an exponent less than $1$ . Taking logs of both variables:

$\log L$	$1.000$	$1.301$	$1.602$	$1.778$	$1.903$	$2.000$
$\log T$	$-0.201$	$-0.046$	$0.104$	$0.190$	$0.253$	$0.303$

x-axis: log(length). y-axis: log(period). The log-log plot is linear.Open in grapher →

The log-log plot is linear, and technology gives $\widehat{\log T} = -0.701 + 0.502 \log L$ with $r^2 \approx 0.9999$ .

Worked example: Predicting from a power model

Predict the period of a $50$ cm pendulum.

\widehat{\log T} = -0.701 + 0.502 \log 50 = -0.701 + 0.502(1.699) = -0.701 + 0.853 = 0.152

\hat{T} = 10^{0.152} \approx 1.42 \text{ seconds.}

A slope of about $0.5$ in the log-log model means $T \approx a \cdot L^{0.5} = a\sqrt{L}$ , which matches the physics: the period is proportional to the square root of the length.

Worked example: Choosing between models

A student fits three models to the same data. Which should she use?

Model	$r^2$	Residual plot
$y$ on $x$	$0.93$	clear curve
$\log y$ on $x$	$0.97$	random scatter
$\log y$ on $\log x$	$0.95$	clear curve

She should use the $\log y$ on $x$ (exponential) model. It is the only one whose residual plot has no leftover pattern, and it also has the largest $r^2$ . If a model with a slightly higher $r^2$ had a curved residual plot, the residual plot would still win: a curved residual plot means the model has the wrong form.

Tip

Before transforming, ask what kind of growth makes sense. Constant amounts per step suggest linear. Constant factors per step suggest exponential ( $\log y$ ). Doubling $x$ multiplies $y$ by a constant factor suggests a power model ( $\log y$ and $\log x$ ).

Practice

Practice 1

Use the bacteria model $\widehat{\log y} = 1.306 + 0.174x$ to predict the count (in thousands) at $x = 6$ hours. Round to the nearest whole number.