Exponential and Logarithmic Data Modeling and Semi-log Plots

📈 Exponential and Logarithmic Data Modeling and Semi-log Plots

Part 1 of 7 — Exponential Models from Two Points and Ratios

Topic Overview

When Is a Data Set Exponential?

A function is exponential, $f(x) = ab^x$, exactly when equal-length input intervals produce outputs in a constant ratio. In a table with equally spaced inputs, divide each output by the one before it: if every ratio is (about) the same number $b$, the data are exponential.

The ratios hover around $1.5$, so an exponential model is reasonable. The differences ($1.1, 1.3, 2.4, 3.3, 5.1$) are clearly not constant, so a linear model is not.

🔍 Building the Model from Two Points

Pick two data points $(x_1, y_1)$ and $(x_2, y_2)$. Because $\dfrac{y_2}{y_1} = \dfrac{ab^{x_2}}{ab^{x_1}} = b^{\,x_2 - x_1}$,

$\boxed{b = \left(\frac{y_2}{y_1}\right)^{\frac{1}{x_2 - x_1}}} \qquad \boxed{a = \frac{y_1}{b^{x_1}}}$

Worked Example 1 — the data above

Use $(0, 2.0)$ and $(4, 10.1)$: $b = \left(\dfrac{10.1}{2.0}\right)^{1/4} = 5.05^{0.25} \approx 1.50$, and $a = 2.0$ because $x_1 = 0$.

Model: $P(t) = 2.0(1.5)^t$ — the population grows about $50\%$ per year. Figure 1 shows the curve threading the points.

Worked Example 2 — neither point at $x = 0$

Find the exponential function through $(1, 10)$ and $(4, 80)$.

$b^{4-1} = \dfrac{80}{10} = 8$, so $b = 2$. Then $a = \dfrac{10}{2^1} = 5$. Model: $f(x) = 5 \cdot 2^x$. Check: $f(4) = 5 \cdot 16 = 80$ ✔

Worked Example 3 — inputs spaced by more than $1$

Every $2$ units the output is multiplied by $0.9$, so $y = 50(0.9)^{x/2}$. Per one unit the factor is $0.9^{1/2} \approx 0.949$: $y \approx 50(0.949)^x$. Both forms are the same function (CED 2.4: $b^{x/2} = (b^{1/2})^x$).

⚠️ Always divide the exponent by the spacing of the inputs, or take the matching root of the ratio. Writing $50(0.9)^x$ would make the quantity shrink twice as fast as the data show.

🗣️ Building the Model from a Description

Reading the Parameters Back

For $f(x) = ab^x$ with $b > 0$:

• $a = f(0)$ is the initial value (the output when the input is $0$).
• $b$ is the ratio between outputs one input unit apart; $b - 1$ is the percent change per unit, written as a decimal.
• The output after $n$ units is the initial value multiplied by $b$ exactly $n$ times.

Justify (AP language): "The data are modeled by an exponential function because over equal-length input intervals the output values change by a constant ratio of approximately $1.5$."

Model Construction Check 🎯

Two-Point Drill 🧮

1) An exponential function passes through $(0, 7)$ and $(2, 28)$. What is $b$? (e.g., through $(0, 5)$ and $(2, 45)$: $b = \sqrt{45/5} = 3$)

2) An exponential function passes through $(1, 6)$ and $(3, 24)$. What is $a$? (e.g., through $(1, 10)$ and $(3, 40)$: $b = 2$, so $a = 10/2 = 5$)

3) A quantity is $100$ at $x = 0$ and $160$ at $x = 2$. Find the growth factor $b$ per unit, rounded to three decimals.

Exponential Model Concepts 🔽

Exit Quiz ✅

🧮 Exponential Regression on the Calculator

Part 2 of 7 — Exponential Regression on the Calculator

Real data rarely give exactly constant ratios. When the ratios are only approximately constant, a graphing calculator's exponential regression finds the single model $y = ab^x$ that best fits all the points at once.

What the Calculator Does for You

On the AP exam you are never asked to compute a regression by hand. Part B of the multiple-choice section and Questions 1–2 of the free-response section allow a graphing calculator; you run the regression, then interpret $a$ and $b$ and use the model. Keep every decimal the calculator shows while you work, and round only the final answer (usually to three decimal places).

Worked Example — Yearly Sales

The ratios are $1.18, 1.23, 1.16, 1.22, 1.17$ — close to constant, so an exponential model is appropriate. ExpReg returns

$a \approx 11.98, \qquad b \approx 1.192 \qquad \Longrightarrow \qquad S(t) = 11.98(1.192)^t$

🗣️ Interpreting and Using the Output

Interpret $a$ and $b$ in context

• $a \approx 11.98$: the model's sales in $2020$ ($t = 0$) are about $11{,}980$ units. (This is the model's value, not the recorded $12.0$ — regression does not have to pass through any data point.)
• $b \approx 1.192$: each year, sales are multiplied by about $1.192$, an increase of about $19.2\%$ per year.

Predict

• $S(7) = 11.98(1.192)^7 \approx 41.0$ thousand units in $2027$.
• $S(2.5) = 11.98(1.192)^{2.5} \approx 18.6$ thousand units halfway through $2022$.

Interpolation vs. Extrapolation

Units matter. Say "about $41.0$ thousand units," not "$41.0$." A model is only as meaningful as the units of its inputs and outputs.

When the model is "good enough"

The regression always produces some exponential curve. Whether that curve is an appropriate model is a separate question, answered by the residual plot in Part 3.

Regression Output Check 🎯

Use the Model 🧮

1) A regression gives $y = 3.2(1.25)^x$. Find $y$ when $x = 3$. (e.g., $y = 2(1.5)^2 = 4.5$)

2) A regression gives $b = 1.072$. By what percent does the model grow per unit of input? Enter the number only. (e.g., $b = 1.03$ means $3$)

3) A regression gives $y = 80(0.9)^x$. Find $y$ when $x = 4$, rounded to one decimal place.

Reading Regression Output 🔽

Exit Quiz ✅

📉 Residuals and Residual Plots

Part 3 of 7 — Residuals and Residual Plots

A model is only useful if it is appropriate for the data. The tool for checking is the residual.

$\boxed{\text{residual} = \text{actual value} - \text{predicted value}}$

Worked Example — the population model from Part 1

Model: $P(t) = 2(1.5)^t$.

At $t = 2$ the residual is $4.4 - 4.5 = -0.1$: the model overestimates the population by $0.1$ thousand. The residuals are small and alternate in sign with no trend.

📊 Reading a Residual Plot

A residual plot graphs each residual against its input. Read it with one question in mind: is there a pattern?

Where Plot B Comes From

Fit a line to the exponential data $(0, 2), (1, 6), (2, 18), (3, 54)$. Linear regression gives $y = 16.8x - 5.2$. Its residuals:

Positive, negative, negative, positive — a U-shape. The line is too low at the ends and too high in the middle because the data curve upward. The exponential model $y = 2 \cdot 3^x$ fits these points exactly, with every residual $0$.

Key fact: the residuals of any regression fit add to about $0$ no matter how bad the model is, so "the residuals are small on average" proves nothing. Only the pattern (or its absence) tells you whether the model is appropriate.

Residual Check 🎯

Compute the Residual 🧮

1) Actual value $7.5$, predicted value $8.1$. Residual $=$ ? (e.g., actual $10$, predicted $9$: residual $= 10 - 9 = 1$)

2) The model is $y = 4(1.2)^x$ and the actual value at $x = 2$ is $6.0$. Residual $=$ ? (e.g., model $2(1.5)^x$, actual $4.4$ at $x = 2$: $4.4 - 4.5 = -0.1$)

3) The model is $y = 16.8x - 5.2$ and the actual value at $x = 2$ is $18$. Residual $=$ ?

Residual Concepts 🔽

Exit Quiz ✅

⚖️ Choosing and Validating a Model

Part 4 of 7 — Choosing and Validating a Model

Given a table of data, which type of function should model it? Compare the pattern of change over equal-length input intervals.

Justify with the pattern, not the picture. "The data are modeled by a quadratic function because the second differences of the output values are constant" is the sentence the exam wants.

Worked Example — Which Model?

• First differences: $3, 4, 5, 7$ — not constant, so not linear.
• Second differences: $1, 1, 2$ — not constant, so a quadratic is doubtful.
• Ratios: $1.30, 1.31, 1.29, 1.32$ — nearly constant, so an exponential model is the best candidate.

Exponential regression gives $a \approx 9.98$ and $b \approx 1.304$: $y \approx 9.98(1.304)^t$.

✅ Validating with Residuals

Compare the exponential model with a linear regression ($y = 4.7t + 8.8$) on the same data.

• Exponential residuals: tiny and alternating in sign, no pattern → appropriate.
• Linear residuals: positive at both ends and negative in the middle, a U-shape → the line misses the upward bend; not appropriate.

Using Context, Too

The numbers choose between candidate models; the context rules models in or out before you even look.

Assumptions and Limitations

Every model assumes the observed pattern continues. State the limits:

• "The exponential model assumes the quantity keeps growing by about $30\%$ per unit; it is reasonable within $0 \leq t \leq 4$ and increasingly uncertain beyond the data."
• Physical limits (a population cannot exceed its habitat; a percent cannot exceed $100$) often make a model valid only on a restricted domain.

Model Selection Check 🎯

Pattern Drill 🧮

1) For the outputs $4, 7, 12, 19, 28$ at inputs $0, 1, 2, 3, 4$, what is the (constant) second difference? (e.g., $1, 4, 9, 16$: differences $3, 5, 7$, second differences $2$)

2) For the outputs $3, 6, 12, 24$ at inputs $0, 1, 2, 3$, what is the common ratio?

3) The line $y = 4.7t + 8.8$ is fit to data with the actual value $17$ at $t = 2$. What is the residual there?

Which Model? 🔽

Exit Quiz ✅

🪵 Logarithmic Functions in Context

Part 5 of 7 — Logarithmic Functions in Context

A logarithmic model has the form

$\boxed{y = a + b\log_c x} \qquad (\text{often } y = a + b\ln x \text{ or } y = a + b\log x)$

Its defining pattern is the mirror image of the exponential pattern:

Multiplying the input by $c$ adds $b$ to the output. Multiplying the input by $c^2$ adds $2b$, and so on.

A Log Model from Data

Each time $x$ is multiplied by $10$, $y$ increases by $3$, so $y = 4 + 3\log x$. Check: $x = 100$ gives $4 + 3(2) = 10$ ✔.

Each doubling of $x$ adds $3$: $y = 2 + 3\log_2 x$ (at $x = 2$: $2 + 3(1) = 5$ ✔; at $x = 16$: $2 + 3(4) = 14$ ✔).

Justify (AP language): "A logarithmic model is appropriate because the output values increase by a constant amount each time the input values are multiplied by the same factor."

📈 Shape, Calculator, and Inverse

What the graph looks like

For $b > 0$, $y = a + b\ln x$ is increasing and concave down: the output rises quickly for small $x$ and then more and more slowly — but it never levels off. There is no horizontal asymptote; $\lim_{x \to \infty} (a + b\ln x) = \infty$. The vertical asymptote is $x = 0$, and the domain is $x > 0$.

Typical contexts: the time to learn a skill as practice hours grow, a reading on a logarithmic instrument scale, the loudness we perceive as sound intensity grows.

Logarithmic regression on the calculator

LnReg fits $y = a + b\ln x$ to a data set and returns $a$ and $b$. You read them off and use the model exactly as in Part 2.

Example: LnReg returns $a = 12.4$ and $b = 5.3$, so $y = 12.4 + 5.3\ln x$. Predicted value at $x = 20$: $12.4 + 5.3\ln 20 \approx 12.4 + 5.3(2.996) \approx 28.3$.

A log model is the inverse of an exponential model

If $y = 2 + 3\log x$, then $\log x = \dfrac{y - 2}{3}$ and $x = 10^{(y-2)/3}$: solving a logarithmic model for the input produces an exponential expression. Swapping the roles of input and output turns a log model into an exponential one and back again (CED 2.10).

Logarithmic Model Check 🎯

Logarithmic Model Drill 🧮

1) For $y = 4 + 3\log x$, find $y$ when $x = 100$. (e.g., at $x = 10$: $4 + 3(1) = 7$)

2) For $y = 2 + 3\log_2 x$, by how much does $y$ increase when $x$ goes from $4$ to $32$? (e.g., from $4$ to $8$: $x$ doubles once, so $y$ increases by $3$)

3) For $y = 4 + 3\log x$, find $x$ when $y = 16$. (e.g., $y = 10$: $\log x = 2$, so $x = 100$)

Logarithmic Model Concepts 🔽

Exit Quiz ✅

📐 Semi-log Plots

Part 6 of 7 — Semi-log Plots

A semi-log plot uses a logarithmic scale on the vertical axis and an ordinary (linear) scale on the horizontal axis. On the vertical axis the labels $1, 10, 100, 1000$ are equally spaced: each step up multiplies the value by $10$. Equivalently, you can plot $\log y$ (or $\ln y$) against $x$ on ordinary axes.

Look at the figure. On linear axes the exponential data $y = 2 \cdot 3^x$ bend sharply upward; on the semi-log axes the same points lie on a straight line.

Why Exponential Data Become Linear

Take the logarithm of both sides of $y = a \cdot b^x$:

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

With $Y = \log y$ this is $Y = (\log b)\,x + \log a$: a line in $x$ and $Y$ with

🔑 Key fact (CED 2.15): data that are linear on a semi-log plot are exponential on ordinary axes, and conversely. A straight semi-log plot is the fingerprint of exponential change; a positive slope means growth ($b > 1$) and a negative slope means decay ($0 < b < 1$).

🔍 Reading the Model from the Line

Worked Example 1

On a plot of $\log y$ against $x$, the data lie on the line through $(0, 1.5)$ and $(4, 2.3)$. Find the exponential model.

• Slope: $\dfrac{2.3 - 1.5}{4 - 0} = 0.2$, so $b = 10^{0.2} \approx 1.585$.
• Intercept: $1.5$, so $a = 10^{1.5} \approx 31.6$.
• Model: $y \approx 31.6(1.585)^x$ — growth of about $58.5\%$ per unit.

Worked Example 2 — neither point on the vertical axis

The semi-log line passes through $(1, 2.0)$ and $(3, 1.0)$.

• Slope: $\dfrac{1.0 - 2.0}{3 - 1} = -0.5$, so $b = 10^{-0.5} \approx 0.316$ (decay).
• Intercept: stepping back one unit from $(1, 2.0)$ adds $0.5$, so $\log a = 2.5$ and $a = 10^{2.5} \approx 316$.
• Model: $y \approx 316(0.316)^x$.

With natural logs

If the vertical axis shows $\ln y$ and the line is $\ln y = 0.3x + 1$, then $a = e^{1} \approx 2.72$ and $b = e^{0.3} \approx 1.35$. The base of the logarithm used on the axis is the base you exponentiate.

Checking the data in Figure form

For $y = 2 \cdot 3^x$: $\log y = 0.301, 0.778, 1.255, 1.732$ at $x = 0, 1, 2, 3$. The $\log y$ values rise by a constant $0.477 = \log 3$ each step (the slope), and the intercept $0.301 = \log 2$ ✔.

Semi-log Check 🎯

Read a Semi-log Line 🧮

1) A semi-log fit gives $\log y = 0.5x + 2$. What is $a = 10^{\text{intercept}}$? (e.g., intercept $1$: $a = 10$)

2) Same fit: what is the base $b = 10^{\text{slope}}$? Use $10^{0.5} \approx 3.16$ and round to two decimals.

3) A natural-log fit gives $\ln y = 2x + 1$. What is the base $b = e^{\text{slope}}$, rounded to two decimals? (e.g., slope $1$: $b = e \approx 2.72$)

Interpret the Semi-log Line 🔽

A data set is plotted as $\log y$ versus $x$ and fits the line $\log y = -0.25x + 3$.

Exit Quiz ✅

🏆 Modeling Synthesis — FRQ Question 2 Skills

Part 7 of 7 — Modeling Synthesis (FRQ Question 2 Skills)

Free-response Question 2 ("Modeling a Non-Periodic Context", calculator allowed) hands you a table or a description, asks for a model, a prediction, an interpretation, and often a judgment about the model. This part walks one such problem end to end.

Your Modeling Toolkit

📋 Worked FRQ-Style Problem

The number $N$ of households in a town with rooftop solar panels was recorded each year. Let $t$ be the number of years since 2018.

$t$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$N$	$40$	$52$	$70$	$91$	$118$	$155$	$200$

(a) Use exponential regression to write a model $N(t) = ab^t$.

The ratios $1.30, 1.35, 1.30, 1.30, 1.31, 1.29$ are nearly constant, so an exponential model is appropriate. ExpReg gives $a \approx 40.22$ and $b \approx 1.309$:

$N(t) = 40.22(1.309)^t$

Interpretation: the model's count in 2018 is about $40$ households, and the number of solar households grows by about $30.9\%$ per year.

(b) Predict the number of solar households in 2026.

$t = 8$: $N(8) = 40.22(1.309)^8 \approx 347$ households. This is an extrapolation two years past the data, so it assumes the $30.9\%$ yearly growth continues.

(c) Find the residual at $t = 3$ and interpret it.

Predicted: $N(3) = 40.22(1.309)^3 \approx 90.2$. Residual $= 91 - 90.2 = +0.8$: the model underestimates the 2021 count by about $1$ household.

(d) Average rate of change from $t = 2$ to $t = 5$.

$\dfrac{N(5) - N(2)}{5 - 2} = \dfrac{155 - 70}{3} \approx 28.3$ households per year (using the data). Because the model is exponential, the average rate of change over later intervals is larger.

(e) On a semi-log plot of $\log N$ against $t$, describe the data.

The points lie approximately on a line with slope $\log 1.309 \approx 0.117$ and vertical intercept $\log 40.22 \approx 1.60$ — the linear pattern on semi-log axes confirms the exponential model.

(f) State a limitation of the model.

The town has a finite number of households, so the model cannot remain valid indefinitely; growth must eventually slow as the remaining households without panels run out. The model is reasonable for $0 \leq t \leq 6$ and nearby years only.

Synthesis Quiz 🎯 (solar-panel data: $N(2) = 70$, $N(5) = 155$; model $N(t) = 40.22(1.309)^t$)

Model Calculations 🧮 (model $N(t) = 40.22(1.309)^t$)

1) Predict $N(8)$, rounded to the nearest whole number. (e.g., $N(7) = 40.22(1.309)^7 \approx 265$)

2) The actual count at $t = 3$ is $91$ and the model gives $90.2$. What is the residual, to one decimal place?

3) On a semi-log plot of $\log N$ against $t$, what is the slope of the line, rounded to three decimals? (e.g., for $b = 2$, the slope is $\log 2 \approx 0.301$)

Modeling Decision Map 🔽

Final Exit Quiz — Data Modeling ✅