AP®MathsCollege BoardPrecalculusStudy GuidesExponential & Logarithmic FunctionsModeling with Exponential & Logarithmic FunctionsSemi-log Plots

Semi-log Plots (College Board AP® Precalculus): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 13 April 2026

Semi-log plots

What is a semi-log plot?

A semi-log plot is a graph in which one of the axes uses a logarithmic scale
- instead of a standard (linear) scale
On a logarithmically scaled axis, each unit of distance represents a multiplicative change rather than an additive one
- E.g. on a standard y-axis, the marks might be at 0, 50, 100, 150, 200, ...
- On a log-scaled y-axis (base 10), the marks are at 1, 10, 100, 1000, ...
  - each step up represents multiplication by 10
In this course, the semi-log plots you encounter will have the y-axis logarithmically scaled and the x-axis on a standard scale

Why do exponential functions appear linear on a semi-log plot?

An exponential function $y = a b^{x}$ has outputs that grow by a constant multiplicative factor ( $b$ ) for each unit increase in $x$
- On a logarithmically scaled y-axis, equal multiplicative changes correspond to equal vertical distances
- So an exponential function, which involves repeated multiplication, produces equally spaced points vertically
  - i.e. the data appears linear on the semi-log plot

One key rule to remember:
- If data points appear to lie along a straight line on a semi-log plot (with logarithmic y-axis)
- then the data can be modeled by an exponential function

E.g. consider the data in the following table

$x$	1	2	3	4	5
$y$	25	45	81	146	262

Note that the ratio of successive terms (45/25, 81/45, etc.) is approximately 1.8 in each case
- This should already let you know that an exponential model would be appropriate!
On a standard plot, these points form a steeply curving exponential curve

Graph with data points at (1, 25), (2, 45), (3, 81), (4, 146), (5, 262) on a grid with axes from 0 to 6 and 0 to 300. — **Data from the table plotted on linearly-scaled axes**

On a semi-log plot (log-scaled y-axis), the same points appear to lie along a straight line

Semilogarithmic graph with data points at (1, 25), (2, 45), (3, 81), (4, 146), (5, 262) on the y-axis with a scale from 1 to 1000. — **Data from the table plotted with logarithmically-scaled y-axis**

Note the uneven spacing of the horizontal grid lines on the semi-log graph
- The lines between 1 and 10 (on the vertical axis) correspond to $y = 2, 3, 4, . . ., 9$
- The lines between 10 and 100 correspond to $y = 20, 30, 40, . . ., 90$
- The lines between 100 and 1000 correspond to $y = 200, 300, 400, . . ., 900$
This sort of unevenly spaced grid is a key sign that a graph is a semi-log plot

Examiner Tips and Tricks

When reading a semi-log plot on the exam, pay close attention to the y-axis scale.

Look for labels like 1, 10, 100, 1000 (or tick marks at powers of 10) to confirm it is logarithmically scaled, rather than a standard scale with evenly spaced numbers

What is the advantage of a semi-log plot over checking ratios?

You can test for exponential behavior by checking whether the ratios of successive output values are constant
- However, this ratio test requires an additive constant to be removed first if the data has a vertical shift
  - e.g. if $y = a b^{x} + k$
A semi-log plot has the advantage of allowing exponential behavior to be spotted
- without having to adjust dependent variable values by a constant first
Specifically, if
- for large input values of a data set
- the logarithms of the output values trend linear
  - i.e. if the output values lie (approximately) along a straight line on a semi-log plot
- then transformations of an exponential function can be used to model the data
This works because for a shifted exponential like $y = a b^{x} + k$
- the exponential term $a b^{x}$ dominates for large $x$
- so the additive constant $k$ becomes negligible
The logarithms of the output values will therefore trend toward a linear pattern as $x$ increases
- even without removing $k$ first
This makes semi-log plots a more robust visual test for exponential behavior than the ratio method

Examiner Tips and Tricks

On the exam, "the data appears linear on a semi-log plot" is strong evidence for an exponential model. But note that the converse is also useful.

I.e., if the data does not appear linear on a semi-log plot, then an exponential model may not be the best fit

Linearising exponential data

How does the linearization work mathematically?

For an exponential model $y = a b^{x}$
- taking the logarithm (base $n$ ) of both sides
  - and using properties of logarithms
- gives

$\begin{array}{rcl} \log_{n} y & = & \log_{n} (a b^{x}) \\ = & \log_{n} a + \log_{n} (b^{x}) \\ = & \log_{n} a + x \log_{n} b \end{array}$

This has the form $Y = c + m x$ , where
- $Y = \log_{n} y$ (the logarithm of the output values)
- $m = \log_{n} b$ (the slope)
  - this is the linear rate of change on the semi-log plot
- $c = \log_{n} a$ (the y-intercept)
  - this is the initial linear value on the semi-log plot
So on a semi-log plot, the linearized model corresponding to $y = a b^{x}$ is
- $\log_{n} y = \log_{n} a + (\log_{n} b) x$

E.g., for $y = 2 \cdot 3^{x}$ (and using base 10 logarithms)
- $\log_{10} y = \log_{10} 2 + (\log_{10} 3) x$
  - or rounding the logs on the right-hand side to 3 decimal places gives
    - $\log_{10} y = 0.301 + 0.477 x$
- The slope of the semi-log plot is $\log_{10} 3 \approx 0.477$
  - and the y-intercept is $\log_{10} 2 \approx 0.301$

How can I use the linearization to build an exponential model?

Because the linearized data follows a straight line
- you can use linear techniques
  - e.g. finding slope and intercept from two points, or linear regression
- to model the data on the semi-log plot
Once you have the slope $m$ and y-intercept $c$ of the line on the semi-log plot
- The base of the exponential function is $b = n^{m}$
  - where $n$ is the logarithm base used
  - This follows from $m = \log_{n} b$
- The initial value is $a = n^{c}$
  - This follows from $c = \log_{n} a$
- The exponential model is $y = a b^{x}$

E.g. suppose on a semi-log plot (base 10), data appears linear with slope $0.3$ and y-intercept 1.5
- Then $b = 10^{0.3} = 1.995262 . . . \approx 2$
  - and $a = 10^{1.5} = 31.622776 . . . \approx 31.623$
- The exponential model is $y \approx 31.623 \cdot 2^{x}$

Examiner Tips and Tricks

The linearization formula $\log_{n} y = (\log_{n} b) x + \log_{n} a$ works with any logarithm base $n$ (as long as $n > 0$ and $n \neq 1$ ).

The most common choice is $n = 10$ (common logarithm), but $n = e$ (natural logarithm) also works

Worked Example

A population of bacteria is measured at regular intervals. The table gives the population $P (t)$ , in thousands, at time $t$ hours.

$t$	0	1	2	3	4
$P (t)$	5	15	45	135	405

(a) Construct the linearization of the data by computing $\log_{10} (P (t))$ for each value in the table.

Answer:

Compute $\log_{10} (P (t))$ for each value:

$\log_{10} 5 = 0.698970 . . . \log_{10} 15 = 1.176091 . . . \log_{10} 45 = 1.653212 . . .$

$\log_{10} 135 = 2.130333 . . . \log_{10} 405 = 2.607455 . . .$

Note that the logarithms increase by $0.477 . . .$ each time, confirming the linear nature of the transformed data

Round to 3 decimal places for your final answers

$t$	0	1	2	3	4
$P (t)$	5	15	45	135	405
$\log_{10} (P (t))$	0.699	1.176	1.653	2.130	2.607

(b) Using the linearized data, find the slope and y-intercept of the linear model on a semi-log plot, and use these to write an exponential model for $P (t)$ .

Answer:

The linearized model is $\log_{10} (P) = c + m t$ , where

the Y-intercept $c$ occurs when $t = 0$

$c = 0.698970 . . .$

the slope between two points can be calculated

$m = \frac{1.176091 . . . - 0.698970 . . .}{1 - 0} = \frac{0.477121 . . .}{1} = 0.477121 . . .$

Convert these to exponential form

$b = 10^{0.477121 . . .} = 3$

$a = 10^{0.698970 . . .} = 5$

So the exponential model is

$P (t) = 5 \cdot 3^{t}$

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Semi-log Plots (College Board AP® Precalculus): Study Guide

Semi-log plots

What is a semi-log plot?

Why do exponential functions appear linear on a semi-log plot?

Examiner Tips and Tricks

What is the advantage of a semi-log plot over checking ratios?

Examiner Tips and Tricks

Linearising exponential data

How does the linearization work mathematically?

How can I use the linearization to build an exponential model?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis