Linearizing Data with Logarithms

Curved data is hard to judge by eye: is it exponential, or a power law like $y = ax^2$? Logarithms solve this problem. Taking logs turns both kinds of curve into straight lines, and straight lines are easy to recognize and easy to fit. From the gradient and intercept of the best-fit line you can read off the parameters of the original model. Logs also let you show data that spans many powers of $10$ on one sensible scale.

Key ideas

Scaling with logarithms

When values range from tiny to huge (masses of animals, populations of towns and countries, website visits), a normal axis squashes most of the data into a corner. Plotting $\log x$ instead of $x$ fixes this: each power of $10$ becomes one equal step, as on the logarithmic scales for pH and decibels. A logarithmic scale is also useful when what matters is the rate of growth rather than the actual size.

Exponential data: the semi-log graph

Suppose $y = ka^x$. Take natural logs and use the laws of logarithms:

$$\ln y = \ln\left(ka^x\right) = \ln k + x\ln a$$

This has the form $Y = mX + c$ with $Y = \ln y$ and $X = x$. So:

• if $y = ka^x$, the graph of $\ln y$ against $x$ is a straight line;
• its gradient is $m = \ln a$, so $a = e^m$;
• its intercept is $c = \ln k$, so $k = e^c$.

A graph with $\ln y$ (or $\log y$) on the vertical axis and plain $x$ on the horizontal axis is called a semi-log graph.

Power data: the log-log graph

Suppose $y = ax^n$. Taking natural logs:

$$\ln y = \ln\left(ax^n\right) = \ln a + n\ln x$$

Now $Y = \ln y$ and $X = \ln x$. So:

• if $y = ax^n$, the graph of $\ln y$ against $\ln x$ is a straight line;
• its gradient is $n$ (the power itself);
• its intercept is $c = \ln a$, so $a = e^c$.

A graph with logs on both axes is called a log-log graph.

Model	Plot	Gradient	Intercept
$y = ka^x$ (exponential)	$\ln y$ against $x$ (semi-log)	$\ln a$	$\ln k$
$y = ax^n$ (power)	$\ln y$ against $\ln x$ (log-log)	$n$	$\ln a$

You can use base-$10$ logs instead of $\ln$; then undo them with $10^{(\ldots)}$ instead of $e^{(\ldots)}$. Just be consistent.

Deciding which model fits

Transform the data both ways and see which plot is closer to a straight line. A quick numerical check is Pearson’s correlation coefficient $r$ for each transformed data set: the one with $|r|$ closer to $1$ is more linear. Then use your GDC’s linear regression on the transformed data to find the best-fit line, and convert its gradient and intercept back into the parameters of the model.

Two practical points:

• A log-log plot needs $x \gt 0$ and $y \gt 0$, and a semi-log plot needs $y \gt 0$, because $\ln 0$ and the logs of negative numbers are undefined.
• Keep full calculator values for the gradient and intercept, and round only the final parameters, because $e^c$ magnifies small rounding errors in $c$.

Reading graphs with logarithmic axes

Published graphs often use axes whose labels go $1, 10, 100, 1000$ at equal spacing (a logarithmic axis) instead of showing $\ln y$ directly. A straight line on such a graph means the same thing: straight on a semi-log graph means exponential, straight on a log-log graph means a power law. To find the model, read two points off the line in their original units and solve for the parameters. In IB examinations you’ll be asked to interpret these graphs, but not to draw or sketch them.

Worked examples

Example 1: Choosing a scale

The typical masses of five animals are: mouse $0.02$ kg, cat $4$ kg, human $70$ kg, elephant $5000$ kg, blue whale $150\,000$ kg. Explain why a logarithmic scale is better for displaying these masses, and find the value of $\log_{10} m$ for each.

Solution. On an ordinary scale from $0$ to $150\,000$ kg, the mouse, cat and human would all sit on top of each other at the far left; even the elephant would be squeezed near $0$. Taking $\log_{10}$:

Animal	mouse	cat	human	elephant	blue whale
$m$ (kg)	$0.02$	$4$	$70$	$5000$	$150\,000$
$\log_{10} m$	$-1.70$	$0.602$	$1.85$	$3.70$	$5.18$

All five values now fit comfortably between $-2$ and $6$, and each step of $1$ means ”$10$ times heavier”. The whale has a $\log_{10}$ mass about $6.9$ greater than the mouse’s, so it’s about $10^{6.88}$ times heavier: $\dfrac{150\,000}{0.02} = 7\,500\,000$, or $7.5$ million times.

Example 2: Exponential or power? A semi-log fit

A biologist counts the bacteria $N$ in a culture every hour.

$t$ (h)	$0$	$1$	$2$	$3$	$4$	$5$
$N$	$520$	$790$	$1230$	$1880$	$2900$	$4450$

• (a) Use a semi-log transformation to show that the data is very close to exponential.
• (b) Find a model of the form $N = ka^t$.
• (c) Interpret $a$.

Solution.

(a) Find $\ln N$ for each value:

$t$	$0$	$1$	$2$	$3$	$4$	$5$
$\ln N$	$6.254$	$6.672$	$7.115$	$7.539$	$7.972$	$8.401$

The values of $\ln N$ go up by almost exactly the same amount (about $0.43$) each hour, so the points $(t, \ln N)$ lie on a straight line. On a GDC, linear regression of $\ln N$ on $t$ gives $r = 1.00$ (to 3 s.f.; more precisely $0.99998$). For comparison, a log-log fit is not possible with $t = 0$, and without that point it gives $r \approx 0.974$, clearly less linear. So the data is exponential.

On a semi-log graph the exponential data becomes a straight line.

(b) The best-fit line is $\ln N = 0.43028t + 6.24977$ (gradient $m \approx 0.430$, intercept $c \approx 6.25$). Convert back:

$$a = e^{m} = e^{0.43028} \approx 1.54, \qquad k = e^{c} = e^{6.24977} \approx 518$$

So $N = 518(1.54)^t$ (parameters to 3 s.f.).

(c) $a \approx 1.54$ means the number of bacteria is multiplied by about $1.54$ every hour: it grows by about $54\%$ per hour.

Example 3: Kepler’s third law from a log-log fit

The table gives each planet’s average distance from the Sun, $r$ (in astronomical units, where Earth’s distance is $1$), and the time $T$ (in years) it takes to orbit the Sun.

Planet	Mercury	Venus	Earth	Mars	Jupiter	Saturn
$r$ (AU)	$0.387$	$0.723$	$1$	$1.524$	$5.203$	$9.537$
$T$ (years)	$0.241$	$0.615$	$1$	$1.881$	$11.86$	$29.46$

Find a model of the form $T = ar^n$.

Solution. For a power model, plot $\ln T$ against $\ln r$:

Planet	Mercury	Venus	Earth	Mars	Jupiter	Saturn
$\ln r$	$-0.949$	$-0.324$	$0$	$0.421$	$1.649$	$2.255$
$\ln T$	$-1.423$	$-0.486$	$0$	$0.632$	$2.473$	$3.383$

Linear regression of $\ln T$ on $\ln r$ gives gradient $1.4997$ and intercept $0.000280$, with $r = 1.00$ (to 3 s.f.). The points lie on a straight line, so a power model fits.

The planets lie on a straight line of gradient $1.5$ on a log-log graph, so $T$ is a power of $r$.

The gradient is the power: $n \approx 1.50$. The intercept gives $a = e^{0.000280} \approx 1.00$. So

$$T = 1.00\,r^{1.50} \qquad\text{(to 3 s.f.), that is, } T \approx r^{3/2}$$

This is Kepler’s third law: the square of the orbital period is proportional to the cube of the distance ($T^2 = r^3$ in these units).

Example 4: Interpreting given best-fit lines

• (a) For one data set, the best-fit line of $\ln y$ against $x$ is $\ln y = 1.6 - 0.25x$. Find the model for $y$ and interpret it.
• (b) For another data set, the best-fit line of $\ln y$ against $\ln x$ is $\ln y = 3.5 - 2\ln x$. Find the model for $y$.

Solution.

(a) A straight line on a semi-log graph means $y = ka^x$, with $\ln k = 1.6$ and $\ln a = -0.25$:

$$k = e^{1.6} \approx 4.95, \qquad a = e^{-0.25} \approx 0.779$$

So $y = 4.95(0.779)^x$. Since $a \lt 1$, $y$ is decaying: it decreases by about $22.1\%$ (because $1 - 0.7788 = 0.2212$) for each increase of $1$ in $x$.

(b) A straight line on a log-log graph means $y = ax^n$, with $n = -2$ and $\ln a = 3.5$:

$$a = e^{3.5} \approx 33.1 \quad\Rightarrow\quad y = 33.1x^{-2} = \frac{33.1}{x^2}$$

This is an inverse-square law, like those on the direct and inverse variation page.

Common mistakes

Mixing up which graph goes with which model. Exponential $y = ka^x$ is straight on a semi-log graph ($\ln y$ against $x$). Power $y = ax^n$ is straight on a log-log graph ($\ln y$ against $\ln x$). The variable in the exponent ($x$ in $a^x$) stays unlogged.

Using the intercept directly as a parameter. The intercept of the line is $\ln k$ (or $\ln a$), not $k$ itself. You must undo the log: $k = e^c$. For a semi-log graph, the gradient is $\ln a$, so $a = e^m$, but for a log-log graph the gradient is already the power $n$.

Mixing bases. If you took $\log_{10}$ of the data, the parameters are $10^c$ and $10^m$, not $e^c$ and $e^m$. Use the same base throughout.

Rounding the intercept too early. $e^{6.25} \approx 518.0$ but $e^{6.2} \approx 492.7$: a small change in $c$ makes a big change in $k$. Keep full values from your GDC and round only at the end.

Trying to log zero or negative values. $\ln 0$ is undefined, so a point with $x = 0$ can’t go on a log-log graph. Leave such points out (and say so), or use a semi-log graph if the model is exponential.

Reading a logarithmic axis as if it were ordinary. Halfway between $10$ and $100$ on a log axis is about $31.6$ (since $10^{1.5} \approx 31.6$), not $55$. Read values at the labelled powers of $10$, or convert using logs.

Practice

1. (Warm-up) The best-fit line of $\ln y$ against $x$ is $\ln y = 0.4x + 1.2$. Find $k$ and $a$ in the model $y = ka^x$.

Solution

$k = e^{1.2} \approx 3.32$ and $a = e^{0.4} \approx 1.49$, so $y = 3.32(1.49)^x$ (3 s.f.).

2. (Warm-up) The best-fit line of $\ln y$ against $\ln x$ is $\ln y = 2.5\ln x - 0.7$. Find the model for $y$.

Solution

This is a power model with $n = 2.5$ and $a = e^{-0.7} \approx 0.497$, so $y = 0.497x^{2.5}$ (3 s.f.).

3. (Warm-up) Three quantities are $0.000\,04$, $3$ and $25\,000$. Find $\log_{10}$ of each to 3 s.f., and explain why these logs would be easier to show on one axis than the original numbers.

Solution

$\log_{10} 0.000\,04 \approx -4.40$, $\log_{10} 3 \approx 0.477$, $\log_{10} 25\,000 \approx 4.40$.

The original numbers differ by a factor of more than $600$ million, so on an ordinary axis the first two would be indistinguishable from $0$. The logs all lie between $-5$ and $5$, so every value is clearly visible.

4. (Core) The value $V$ dollars of a car $t$ years after it was bought is shown.

$t$ (years)	$0$	$1$	$2$	$3$	$4$	$5$
$V$ ($)	$32\,000$	$27\,100$	$23\,200$	$19\,700$	$16\,800$	$14\,300$

• (a) Find the equation of the best-fit line of $\ln V$ against $t$, and the value of $r$.
• (b) Hence find a model $V = ka^t$, and the annual percentage decrease.
• (c) Use the model to estimate when the car will be worth $10 000.

Solution

(a) The values of $\ln V$ are $10.373$, $10.207$, $10.052$, $9.888$, $9.729$, $9.568$. Linear regression gives

$$\ln V = -0.16073t + 10.37151, \qquad r = -1.00 \ (\text{to 3 s.f.; } -0.99998)$$

(b) $a = e^{-0.16073} \approx 0.852$ and $k = e^{10.37151} \approx 31\,900$. So $V = 31\,900(0.852)^t$. Since $1 - 0.8515 = 0.1485$, the car loses about $14.8\%$ of its value each year.

(c) Solve $\ln V = \ln 10\,000$ using the line: $-0.16073t + 10.37151 = \ln 10\,000 \approx 9.21034$, so

$$t = \frac{10.37151 - 9.21034}{0.16073} \approx 7.22 \text{ years}$$

5. (Core) The period $T$ (s) of a pendulum is measured for different lengths $L$ (m).

$L$ (m)	$0.1$	$0.25$	$0.5$	$1.0$	$2.0$
$T$ (s)	$0.63$	$1.00$	$1.42$	$2.01$	$2.84$

• (a) Use a log-log transformation to find a model $T = aL^n$.
• (b) Find the length of a pendulum with a period of $3$ s.

Solution

(a) Linear regression of $\ln T$ on $\ln L$ gives gradient $0.50289$ and intercept $0.69713$, with $r = 1.00$ (to 3 s.f.). So $n \approx 0.503$ and $a = e^{0.69713} \approx 2.01$:

$$T = 2.01L^{0.503}$$

(This matches the physics formula $T = 2\pi\sqrt{L/g} \approx 2.01L^{0.5}$.)

(b) Solve $2.00799L^{0.50289} = 3$ (using full values): $L = \left(\dfrac{3}{2.00799}\right)^{1/0.50289} \approx 2.22$ m.

6. (Core) A best-fit line of $\log_{10} y$ against $x$ has gradient $-0.15$ and intercept $2.8$. Find the model $y = ka^x$ and describe what $a$ tells you.

Solution

Base $10$, so undo with powers of $10$: $k = 10^{2.8} \approx 631$ and $a = 10^{-0.15} \approx 0.708$. So $y = 631(0.708)^x$.

Since $1 - 0.7079 = 0.2921$, $y$ decreases by about $29.2\%$ each time $x$ increases by $1$.

7. (Core) A graph has an ordinary horizontal axis for $x$ and a logarithmic vertical axis for $y$ (labelled $10$, $100$, $1000$, $10\,000$). The data lies on a straight line through the points $(0, 200)$ and $(10, 3200)$, read in original units. Find the model and interpret it.

Solution

A straight line on a semi-log graph means $y = ka^x$. At $x = 0$, $y = k = 200$. Then

$$200a^{10} = 3200 \quad\Rightarrow\quad a^{10} = 16 \quad\Rightarrow\quad a = 16^{1/10} \approx 1.32$$

So $y = 200(1.32)^x$ (3 s.f.): $y$ grows by about $32\%$ for each increase of $1$ in $x$ (it doubles every $2.5$ units, since $16 = 2^4$ over $10$ units).

8. (Challenge) For the data below, decide whether an exponential or a power model fits better by comparing $r$ for a semi-log and a log-log transformation. Find the better model and use it to predict $y$ when $x = 10$.

$x$	$1$	$2$	$3$	$4$	$6$	$8$
$y$	$3.0$	$10.6$	$21.5$	$36.6$	$74.8$	$127$

Solution

Semi-log ($\ln y$ against $x$): $r \approx 0.955$. Log-log ($\ln y$ against $\ln x$): $r \approx 1.00$ ($0.99998$). The log-log plot is far more linear, so use a power model.

The log-log best-fit line is $\ln y = 1.79710\ln x + 1.10306$, so $n \approx 1.80$ and $a = e^{1.10306} \approx 3.01$:

$$y = 3.01x^{1.80}$$

At $x = 10$: $y = 3.01339 \times 10^{1.79710} \approx 189$ (3 s.f.). This is a small extrapolation beyond $x = 8$, so treat it with some caution.

9. (Challenge) On a log-log graph (both axes logarithmic, in original units), a straight line passes through $(2, 50)$ and $(20, 8)$. Find the model $y = ax^n$.

Solution

Both points satisfy $y = ax^n$, so divide:

$$\frac{8}{50} = \frac{a \cdot 20^n}{a \cdot 2^n} = 10^n \quad\Rightarrow\quad n = \log_{10} 0.16 \approx -0.796$$

Then $a = \dfrac{50}{2^n} = 50 \times 2^{0.79588} \approx 86.8$. So $y = 86.8x^{-0.796}$ (3 s.f.).

Check with the other point: $86.8068 \times 20^{-0.79588} \approx 8.00$. ✓