Semi-log plots — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Linearization of exponential functions, conversion between natural and common log semi-log axes, finding exponential model parameters from plotted data, interpreting slope on semi-log plots, and reconverting linear models back to exponential form.

You should already know: Properties of logarithms, general form of exponential functions, slope-intercept form of linear equations.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Semi-log plots?

A semi-log plot is a graph where one axis uses a linear scale and the other uses a logarithmic scale. For exponential functions, which are the focus of AP Precalculus for this topic, we always use a semi-log plot with a linear horizontal (x) axis and a logarithmic vertical (y) axis. This topic is explicitly named in the AP Precalculus CED for Unit 2, accounting for approximately 2-3% of the total exam score. Semi-log plots appear in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ typically ask for the base of the exponential or the slope of the linearized line, while FRQ ask you to construct an exponential model from raw data using semi-log linearization. Synonyms sometimes used include semi-logarithmic plots, which refer to the same structure. The key purpose of a semi-log plot is to linearize exponential relationships, turning a curved exponential graph into a straight line. This makes it far easier to estimate parameters (the initial value and growth/decay rate) from experimental or real-world data that follows an exponential pattern.

2. Linearizing Exponential Functions on Semi-log Axes

The core idea of semi-log plots is converting a non-linear exponential relationship into a linear one, which is much easier to analyze. Start with the standard form of an exponential function: $$y=a{b}^x$$ where $a>0$ is the initial value when $x=0$, $b>0,b\ne 1$ is the constant growth/decay factor, and $x$ is the independent variable (almost always time in AP problems). To linearize this, we take the logarithm of both sides. This works for any logarithm base, but AP most commonly uses base 10 or natural base $e$. Applying logarithm product and power rules: $$\log (y)=\log (a{b}^x)=\log (a)+x\log (b)$$ If we let $Y=\log (y)$, this rearranges to the familiar slope-intercept form of a line: $$Y=\left(\log b\right)x+\log a$$ On a semi-log plot, the x-axis is linear (for $x$) and the y-axis is logarithmic, which is equivalent to plotting $Y=\log (y)$ on a linear axis. This means any exponential function will graph as a perfectly straight line on a semi-log plot, with slope equal to $\log b$ and y-intercept equal to $\log a$. If using natural logarithm, the form is identical: $\ln y=(\ln b)x+\ln a$, with slope $\ln b$ and intercept $\ln a$.

Worked Example

Problem: Given the exponential model $y=12(1.8{)}^x$, write the equation of the linearized line for a base-10 semi-log plot (with linear x-axis, logarithmic y-axis), and find the y-intercept of the linearized line rounded to two decimal places.

1. Start with the original exponential equation: $y=12(1.8{)}^x$
2. Take base-10 logarithm of both sides, apply logarithm rules: ${\log }_{10}y={\log }_{10}(12)+x{\log }_{10}(1.8)$
3. Let $Y={\log }_{10}y$, so the linear equation is $Y=(\log 1.8)x+\log 12$
4. Calculate the y-intercept value: $\log 12\approx 1.08$, so the y-intercept of the linearized line is at $(0,1.08)$.

Exam tip: On the AP exam, always check if the question specifies base 10 or natural log for the semi-log plot; if it does not specify, either form is acceptable as long as you correctly relate the slope to the base of the exponential.

3. Recovering an Exponential Model From a Semi-log Line

The reverse of linearization, recovering the original exponential model from a straight line on a semi-log plot, is the most commonly tested skill for this topic on the AP exam. The process reverses the steps we used for linearization: we start with the linear equation for $Y=\log y$, then exponentiate both sides with the same base used for the logarithm to get back to $y$. If we have $Y=mx+c$ (where $Y={\log }_{10}y$), exponentiating with base 10 gives: $$10^Y=10^{mx+c}⟹y=10^c\cdot (10^m{)}^x$$ This matches the original exponential form $y=a{b}^x$, so we can directly read off the parameters: $a=10^c$ and $b=10^m$. For natural log, the process is identical: $a=e^c$ and $b=e^m$, where $c$ is the intercept and $m$ is the slope of the linearized line.

Worked Example

Problem: A linearized line on a natural log semi-log plot (linear x, log y) has equation $Y=0.25x+1.386$, where $Y=\ln y$. Find the original exponential model $y=a{b}^x$, rounding $a$ and $b$ to two decimal places.

1. Start with the given linearized equation: $Y=\ln y=0.25x+1.386$
2. Exponentiate both sides with base $e$ to eliminate the natural logarithm: $e^{\ln y}=e^{0.25x+1.386}$
3. Simplify using exponent rules for addition: $y=e^{1.386}\cdot (e^{0.25}{)}^x$
4. Calculate the numerical values of the parameters: $e^{1.386}\approx 4.00$, $e^{0.25}\approx 1.28$, so the exponential model is $y=4.00(1.28{)}^x$.

Exam tip: When recovering the model, always write out the exponentiation step explicitly to avoid swapping the intercept and slope values; it is easy to mix up which parameter corresponds to which term.

4. Interpreting Slope and Intercept in Context

AP Precalculus regularly asks for interpretation of semi-log plot parameters in real-world contexts, so understanding what slope and intercept mean beyond just calculation is critical. The intercept $c=\log a$ corresponds to $\log y$ when $x=0$, so exponentiating gives $a$, the initial value of $y$ when the independent variable $x$ is 0. The slope $m=\log b$ means that a 1-unit increase in $x$ causes an $m$-unit increase in $\log y$, which corresponds to multiplying $y$ by $b=10^m$ (for base 10) or $e^m$ (for natural log). A positive slope means $b>1$, so exponential growth; a negative slope means $0<b<1$, so exponential decay. This interpretation is the basis for many context-based MCQ and FRQ parts.

Worked Example

Problem: A demographer studying population growth plots population data on a base-10 semi-log plot, where $x$ is time in decades and $y$ is total population. The linearized line has a slope of 0.3010. What is the decadal population growth factor?

1. For base-10 semi-log plots, the relationship between slope $m$ and growth factor $b$ is $m={\log }_{10}b$.
2. Substitute the given slope: $0.3010={\log }_{10}b$.
3. Rewrite in exponential form to solve for $b$: $b=10^{0.3010}$.
4. Calculate the value: $10^{0.3010}\approx 2$, so the decadal growth factor is 2, meaning the population doubles every 10 years.

Exam tip: Always match the units of the 1-unit x-increase to the context; if x is in centuries, the growth factor you calculate is per century, not per year.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Swapping the values of $a$ and $b$ by assigning the intercept to the base and slope to the initial value. Why: Students mix up which term goes where when reversing linearization, because both the intercept and slope are logarithms of parameters. Correct move: Always write out the full derivation step-by-step: $\log y=(\log b)x+\log a$, so slope = $\log b$, intercept = $\log a$, before calculating values.
• Wrong move: Using the wrong base when exponentiating, e.g., using base $e$ for a base-10 semi-log plot. Why: Students forget that the base of the logarithm on the y-axis determines the base for exponentiation. Correct move: Circle the base of the log specified in the problem before starting calculations, and always use that base when recovering $a$ and $b$.
• Wrong move: Claiming the y-intercept of the semi-log line is the initial value of the exponential function. Why: Students confuse the linearized y-intercept with the original function's intercept. Correct move: Remember that the y-intercept of the line is $\log a$, so you must exponentiate it to get the actual initial value $a$ of the exponential.
• Wrong move: Linearizing $y=a{b}^x$ by taking the log of $x$ instead of $y$ for a standard semi-log plot. Why: Students confuse semi-log plots (one log axis) with log-log plots (two log axes), used for power functions. Correct move: For exponential functions $y=a{b}^x$, we always linearize by logging the dependent $y$-variable, leaving $x$ linear.
• Wrong move: Trying to take the logarithm of a negative $y$-value to fit a semi-log plot. Why: Students forget that logarithms are only defined for positive inputs. Correct move: Recognize that semi-log plots can only be used for positive $y$-values; any negative $y$-values in the dataset are errors or do not follow an exponential model.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

A semi-log plot (linear x-axis, natural log y-axis) of an exponential function $y=a{b}^x$ has a slope of -0.6931. What is the base $b$ of the exponential function? A) -0.693 B) 0.5 C) 2 D) 0.693

Worked Solution: For a natural log semi-log plot, the slope of the linearized line equals $\ln b$ by definition. Substituting the given slope of -0.6931 gives the equation $\ln b=-0.6931$. Exponentiating both sides with base $e$ gives $b=e^{-0.6931}\approx 0.5$. The common distractors reflect common mistakes: option C uses the wrong sign for the slope, while options A and D mistake the slope itself for the base $b$. The correct answer is B.

Question 2 (Free Response)

The table below gives data for the value of a car over time, where $x$ is years after purchase and $y$ is car value in thousands of dollars:

(a) Complete the table of values for $Y=\ln y$, rounded to two decimal places, then write the equation of the linearized line for a semi-log plot. (b) Use the linearized line to find the exponential decay model $y=a{b}^x$ for the car value. (c) According to your model, what will be the value of the car (in thousands of dollars) 5 years after purchase, rounded to two decimal places?

Worked Solution: (a) Calculate $\ln y$ for each row: $x=0:\ln 32\approx 3.47$, $x=1:\ln 25.6\approx 3.24$, $x=2:\ln 20.48\approx 3.02$, $x=3:\ln 16.384\approx 2.79$. The slope between any two consecutive points is $-0.23$, so the linearized equation is $Y=-0.23x+3.47$. (b) Recover the exponential model by exponentiating both sides: $y=e^{3.47}(e^{-0.23}{)}^x\approx 32(0.79{)}^x$. (c) Substitute $x=5$ into the model: $y=32(0.79{)}^5\approx 32(0.3077)\approx 9.85$ thousand dollars.

Question 3 (Application / Real-World Style)

A physicist measures the intensity of radiation from a radioactive source as a function of distance from the source, and finds that intensity $I$ (in counts per second) follows an exponential relationship with distance $d$ (in meters). The data plotted on a base-10 semi-log plot (linear $d$, log $I$) gives a straight line with slope -0.3010. If the intensity at 1 meter is 120 counts per second, write the model for $I(d)$ and find the intensity at 4 meters.

Worked Solution:

1. For base-10 semi-log, the linearized equation is $\log I=md+c=-0.3010d+c$.
2. Substitute the known point $d=1,I=120$: $\log 120=-0.3010(1)+c$, so $c=\log 120+0.3010\approx 2.079+0.3010=2.38$.
3. Exponentiate to get the final model: $I(d)=10^{2.38}(10^{-0.3010}{)}^d\approx 240(0.5{)}^d$.
4. Substitute $d=4$: $I(4)=240(0.5{)}^4=240\left(\frac{1}{16}\right)=15$ counts per second.

In context, this means the expected radiation intensity 4 meters from the source is 15 counts per second.

7. Quick Reference Cheatsheet

8. What's Next

Semi-log plots are the foundation for linear regression of exponential models, which is the next core topic you will encounter in Unit 2. Without understanding how semi-log linearization works, you will not be able to correctly fit exponential models to real-world data using linear regression, a key skill tested heavily on the AP exam. This topic also connects directly to log-log plots for power functions, which use the same linearization concept for a different family of non-linear functions. Semi-log plots also reinforce the core inverse relationship between exponential and logarithmic functions that is the central theme of Unit 2, and mastery here will simplify all exponential modeling questions on the exam.

• Exponential growth and decay models
• Logarithm properties
• Log-log plots
• Linear regression for non-linear models