Semi-Log Plots: Exponential Data as a Line

A semi-log plot is a graph with a logarithmic scale on the y-axis and a regular scale on the x-axis. Exponential data, $y = a \cdot b^x$ , makes a straight line on it, because $\log y = \log a + x \log b$ is linear in $x$ . The line’s slope is $\log b$ and its intercept is $\log a$ , so you can read the model from the line.

Before this lesson: You should know Logarithmic Scales, Exponential Functions and Properties of Logarithms.

Why Exponential Data Looks Straight

Exponential data grows by the same factor each step. On regular axes, that makes a curve that bends up faster and faster. On a log scale, every equal factor is an equal step up. So equal factors become equal rises, and the points line up.

The algebra shows the same thing. Take the log of both sides of $y = a \cdot b^x$ . The product rule and power rule give $\log y = \log a + x \log b$ . That has the form $Y = mx + c$ with $Y = \log y$ , slope $m = \log b$ and intercept $c = \log a$ .

Two graphs of the same green data points y = 3, 6, 12, 24, 48, 96 for x = 0 to 5. Left, on regular axes, the solid blue curve bends upward. Right, on a semi-log graph whose y-axis is marked 1, 10 and 100 with log spacing, the points fall on a dashed orange straight line. — The same exponential data, y = 3 · 2^x, curves on regular axes but makes a straight line when the y-axis uses a log scale.

Semi-Log Rules

y = a \cdot b^x \quad \Longleftrightarrow \quad \log y = (\log b)x + \log a

If the points $(x, \log y)$ lie on a line, the data is exponential.
Slope of the line: $m = \log b$ , so the growth factor is $b = 10^m$ .
Intercept of the line: $c = \log a$ , so the starting value is $a = 10^c$ .
A positive slope means growth ( $b \gt 1$ ); a negative slope means decay ( $0 \lt b \lt 1$ ).

If the line uses the natural log, $\ln y = mx + c$ , then $b = e^m$ and $a = e^c$ .

Worked Examples

Use a calculator for logs. Round logs to three decimal places.

Example 1: Straighten exponential data

Find $\log y$ for $y = 3 \cdot 2^x$ at $x = 0, 1, 2, 3, 4$ , and show the logs make a line.

The $y$ -values are $3, 6, 12, 24, 48$

0.477, 0.778, 1.079, 1.380, 1.681

Take each log.

Subtract neighbors: $0.778 - 0.477 = 0.301$ , $1.079 - 0.778 = 0.301$ , $1.380 - 1.079 = 0.301$ , $1.681 - 1.380 = 0.301$

The differences are equal, so the points $(x, \log y)$ form a line with slope 0.301

Compare with the rule: $\log 2 \approx 0.301$ , the log of the growth factor

Answer: the logs rise by $\log 2 \approx 0.301$ each step, so they make a straight line, as in the right graph above.

Example 2: Test data for an exponential pattern

Which data set is exponential: A: $5, 15, 45, 135$ or B: $5, 15, 25, 35$ , both at $x = 0, 1, 2, 3$ ?

0.699, 1.176, 1.653, 2.130

Logs for A.

Differences for A: $0.477, 0.477, 0.477$ , all equal to $\log 3$

0.699, 1.176, 1.398, 1.544

Logs for B.

0.477, 0.222, 0.146

Differences for B (not equal).

Answer: A is exponential (it triples each step), so it is straight on a semi-log plot. B is linear, so on a semi-log plot it would curve and flatten out.

Example 3: From a semi-log line to a model

A semi-log line has equation $\log y = 0.2x + 1.5$ . Write the exponential model.

Match the rule: slope $m = 0.2$ and intercept $c = 1.5$

b = 10^{0.2} \approx 1.585

Find the growth factor: 10 to the power of the slope.

a = 10^{1.5} \approx 31.62

Find the starting value: 10 to the power of the intercept.

y \approx 31.62(1.585)^x

Write the model.

Answer: $y \approx 31.62(1.585)^x$ , growing about 58.5% per step.

Example 4: A model from two points

An exponential model passes through $(2, 50)$ and $(6, 800)$ . Use logs to find it.

Take logs of the $y$ -values: $\log 50 \approx 1.699$ and $\log 800 \approx 2.903$

m = \frac{2.903 - 1.699}{6 - 2} = \frac{1.204}{4} = 0.301

Find the slope of the semi-log line (see slope)

b = 10^{0.301} \approx 2

Find the growth factor: 10 to the power of the slope.

\log a = 1.699 - 2(0.301) = 1.097

Find the intercept of the log line.

a = 10^{1.097} \approx 12.5

Undo the log: raise 10 to that power.

Check: $12.5 \cdot 2^2 = 50$ and $12.5 \cdot 2^6 = 800$

Answer: $y = 12.5 \cdot 2^x$

Example 5: Word problem with bacteria

A lab counts bacteria every hour: 200, 340, 578, 983 and 1,670 at $t = 0, 1, 2, 3, 4$ . Show the growth is exponential and find the hourly growth factor.

2.301, 2.531, 2.762, 2.993, 3.223

Take logs.

Differences: $0.230, 0.231, 0.231, 0.230$ , nearly equal, so the semi-log plot is a straight line

m = \frac{3.223 - 2.301}{4 - 0} = \frac{0.922}{4} \approx 0.2305

Find the slope from the first and last points.

b = 10^{0.2305} \approx 1.70

Find the growth factor: 10 to the power of the slope.

Starting value: $a = 200$ , the count at $t = 0$

Answer: $N(t) \approx 200(1.70)^t$ . The population grows about 70% each hour.

Common Mistakes

Mistake	Example of the error	Fix
Using the slope as the growth factor	Slope 0.301, so $b = 0.301$	The slope is $\log b$ . Undo the log: $b = 10^{0.301} \approx 2$ .
Reading the log axis like a regular axis	Thinking the point halfway between 10 and 100 is 55	Halfway on a log axis is $10^{1.5} \approx 31.6$ .
Taking the log of x too	Plotting $(\log x, \log y)$	A semi-log plot only logs $y$ . Logging both is a log-log plot, used for power functions.
Mixing log and ln	Using $b = e^m$ after taking base-10 logs	Match the base: base 10 uses $10^m$ ; natural log uses $e^m$ .
Expecting linear data to look straight	Plotting $5, 15, 25, 35$ on a semi-log graph	Linear data curves and flattens on a semi-log plot.

Related Methods

Exponential regression

With messy real data, the points won’t be exactly on a line. Fitting the best line to $(x, \log y)$ is one way to do exponential regression.

Natural logs

Scientists often use $\ln y$ . Then the slope $k$ gives the model $y = a e^{kx}$ , using the number e. Growth is $k \gt 0$ and decay is $k \lt 0$ .

Log-log plots

If both axes use log scales, power functions like $y = ax^n$ become straight lines with slope $n$ .

Practice Problems

The points $(0, 10)$ , $(1, 100)$ and $(2, 1000)$ are plotted on a semi-log graph. What is the slope of the line, and what is the model?
Is the data $4, 12, 36, 108$ (at $x = 0, 1, 2, 3$ ) exponential? Use logs.
Is the data $2, 4, 6, 8$ (at $x = 0, 1, 2, 3$ ) exponential? Use logs.
A semi-log line is $\log y = 0.5x + 1$ . Write the model.
A semi-log line is $\log y = -0.1x + 2$ . Write the model. Is it growth or decay?
A semi-log line passes through $(0, 1)$ and $(4, 3)$ , in the form $(x, \log y)$ . Write the model.
Use logs to find the exponential model through $(1, 6)$ and $(3, 54)$ .
Find the slope and intercept of the semi-log line for $y = 5(1.08)^x$ .
A natural-log plot gives the line $\ln y = 0.3x + 2$ . Write the model.
An account holds $1,000, $1,060, $1,123.60 and $1,191.02 in years 0 to 3. Use logs to find the yearly growth rate.

Show answers

Logs: $1, 2, 3$ , so the slope is $\frac{2 - 1}{1 - 0} = 1$
Growth factor $10^1 = 10$ , starting value $10^1 = 10$
Answer: slope 1; $y = 10 \cdot 10^x$

0.602, 1.079, 1.556, 2.033

Take the log of each value.

0.477, 0.477, 0.477

Find the differences.

Answer: yes; it triples each step ( $10^{0.477} \approx 3$ )

0.301, 0.602, 0.778, 0.903

Take the log of each value.

0.301, 0.176, 0.125

Find the differences. They are not equal.

Answer: no; it is linear (adds 2 each step)

$b = 10^{0.5} \approx 3.162$ and $a = 10^1 = 10$
Answer: $y \approx 10(3.162)^x$

$b = 10^{-0.1} \approx 0.794$ and $a = 10^2 = 100$
Answer: $y \approx 100(0.794)^x$ , decay (about 20.6% per step)

Slope: $\frac{3 - 1}{4 - 0} = 0.5$ ; intercept: 1
$b = 10^{0.5} \approx 3.162$ , $a = 10^1 = 10$
Answer: $y \approx 10(3.162)^x$

Logs: $\log 6 \approx 0.778$ and $\log 54 \approx 1.732$

\frac{1.732 - 0.778}{3 - 1} = \frac{0.954}{2} = 0.477

Find the slope of the log line.

b = 10^{0.477} \approx 3

Undo the log: raise 10 to the slope.

a = \frac{6}{3} = 2

Find the starting value: divide by the growth factor.

Answer: $y = 2 \cdot 3^x$ (check: $2 \cdot 27 = 54$ )

\log 1.08 \approx 0.0334

The slope is the log of the growth factor.

\log 5 \approx 0.699

The intercept is the log of the starting value.

Answer: $\log y \approx 0.0334x + 0.699$

$a = e^2 \approx 7.389$ and $b = e^{0.3} \approx 1.350$
Answer: $y \approx 7.389e^{0.3x} \approx 7.389(1.350)^x$

10.

Logs: $3, 3.0253, 3.0506, 3.0759$ , differences all about 0.0253

10^{0.0253} \approx 1.06

Find the growth factor: 10 to the power of the slope.

Answer: about 6% per year

FAQ

What is a semi-log plot used for?

It shows whether data is exponential. Exponential data makes a straight line, and the line’s slope and intercept give the growth factor and starting value.

How do you read a log axis?

Each labeled mark is 10 times the one below it. The spacing between 1 and 2 is wider than between 8 and 9, because the axis measures the log of each value.

What does the slope of a semi-log plot mean?

The slope is the log of the growth factor. A slope of $m$ means the quantity multiplies by $10^m$ for each 1-unit step in $x$ .

What is the difference between semi-log and log-log plots?

A semi-log plot uses a log scale on one axis, which straightens exponential functions. A log-log plot uses log scales on both axes, which straightens power functions.

Summary

A semi-log plot uses a log scale on the y-axis only.
Exponential data $y = a \cdot b^x$ becomes the line $\log y = (\log b)x + \log a$ .
Growth factor: $b = 10^{\text{slope}}$ ; starting value: $a = 10^{\text{intercept}}$ .
Equal differences in $\log y$ mean the data is exponential.
For natural logs, use $e^{\text{slope}}$ and $e^{\text{intercept}}$ .

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