Unit 2 · Topic 2.14

Logarithmic Function Context and Data Modeling

Use a logarithmic model y = a + bln x when the input must multiply by a constant for the output to rise by a constant.

Practice 1: Procedural and Symbolic FluencyPractice 3: Communication and Reasoning

Part 1 · Hook

Why this matters

Practice a new skill and the first weeks bring big gains; later the same effort buys less. Doubling your practice time adds about the same improvement each time. Data like that, where the input has to multiply to make the output climb by equal steps, call for a logarithmic model.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $y = a + b\ln x$ at $x = 1$ equals

$a$
$a + b$
$b$
$0$

Show the answer

$\ln 1 = 0$ .

Correct: $a$ :
$a + b$ :
$b$ :
$0$ :

2. A residual is

the actual value minus the predicted value
the predicted value minus the actual value
the predicted value divided by the actual value
the slope of the model

Show the answer

Residual = actual − predicted.

Correct: the actual value minus the predicted value:
the predicted value minus the actual value:
the predicted value divided by the actual value:
the slope of the model:

3. Solve $\ln x = 2.5$ .

$x = e^{2.5}$
$x = 2.5e$
$x = \frac{2.5}{e}$
$x = 10^{2.5}$

Show the answer

Exponential form, base $e$ .

Correct: $x = e^{2.5}$ :
$x = 2.5e$ :
$x = \frac{2.5}{e}$ :
$x = 10^{2.5}$ :

Part 4 · See it

See it first

Each doubling of the weeks adds

12.5

words per minute; the model

y = 52 + 18\ln x

fits the data and rises more and more slowly. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Inputs multiply by a constant factor while outputs add a constantthe data fit $y = a + b\log_c x$ , a logarithmic model
One point has $x = 1$ the logarithm is $0$ there, so the output at $x = 1$ is $a$
Data are not exactuse logarithmic regression, $y = a + b\ln x$ , and store the constants
Solve for an inputisolate the logarithm and rewrite as a power
Move far past the data or near $x = 0$ the model is unsupported or gives impossible values, so restrict the domain

Part 6 · Key ideas

Key ideas

A logarithmic model $y = a + b\ln x$ fits when equal factors in $x$ give equal changes in $y$ .
Build it from two points (one with $x = 1$ makes $a$ easy) or with logarithmic regression on a calculator.
$b > 0$ : increasing, concave down, so the rate of change decreases: growth that slows but never stops.
Justify the choice with the numbers: "each time $x$ triples, $y$ increases by $6$ ".

Part 7 · Misconception

A common mistake

The wrong idea: A logarithmic model levels off at a maximum value, like a population reaching its limit.

What actually happens: A logarithmic model grows without bound: $\lim_{x \to \infty}(a + b\ln x)$ $= \infty$ for $b > 0$ . Its rate of change shrinks, so the growth looks like leveling off, but there is no ceiling.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Typing speed and practice

Typing speed, in words per minute, after x hours of practice
$x$ (hours)	1	3	9	27	81
$y$ (words per minute)	20	26	32	38	44

1. Why is a logarithmic model a good choice for these data?

Each time the hours are multiplied by $3$ , the speed increases by $6$
The speed increases by the same amount over equal-length intervals of hours
The ratios of successive speeds are equal
The second differences of the speeds are equal over equal-length intervals

Show the answer

The inputs grow by a constant factor ( $\times 3$ ) while the outputs grow by a constant amount ( $+6$ ). That is the pattern of $y = a + b\log_3 x$ .

Correct: Each time the hours are multiplied by $3$ , the speed increases by $6$ : Multiplying the input adds a fixed amount to the output: logarithmic.
The speed increases by the same amount over equal-length intervals of hours: The input intervals are not equal: $2, 6, 18, 54$ hours. Equal differences over equal intervals would be linear.
The ratios of successive speeds are equal: $\frac{26}{20} = 1.3$ but $\frac{32}{26} \approx 1.23$ : the ratios are not equal, so not exponential.
The second differences of the speeds are equal over equal-length intervals: The inputs are not equally spaced, so differences of outputs cannot test a quadratic here.

2. Write a model $y = a + b\log_3 x$ that fits every point in the table.

Type an expression for $y =$ .

Show the answer

At $x = 1$ , $\log_3 1 = 0$ , so $a = 20$ . Each factor of $3$ adds $1$ to $\log_3 x$ and $6$ to $y$ , so $b = 6$ : $y = 20 + 6\log_3 x$ .

Answer: $y =$ $20 + 6 \cdot \log_{3}(x)$

3. Use the model $y = 20 + 6\log_3 x$ to predict the typing speed after $243$ hours.

No calculator. Type the exact value (a fraction, π, e or a root is fine), in words per minute.

Show the answer

$243 = 3^5$ , so $y = 20 + 6 \cdot 5 = 50$ words per minute. It continues the pattern: one more factor of $3$ after $81$ adds $6$ .

Answer: 50 words per minute

Table

Height of a sapling

A logarithmic regression

H = a + b\ln x

models the data. Use a graphing calculator and store the constants.

Height of a tree sapling, in centimeters, at selected ages, in months
$x$ (months)	1	2	4	6	10	15
$H$ (centimeters)	12.0	21.9	30.0	35.2	43.5	48.4

4. Find $b$ in the logarithmic regression $H = a + b\ln x$ for the data. Give it correct to three decimal places.

Calculator allowed. Give the answer correct to three decimal places.

Show the answer

Enter $x$ and $H$ as lists and run a logarithmic regression ( $y = a + b\ln x$ ): $a \approx 12.000$ and $b \approx 13.398$ . Store both for later parts.

Answer: 13.398

5. Use the regression model to predict the sapling's height at $12$ months. Give it correct to three decimal places.

Calculator allowed. Give the answer correct to three decimal places, in centimeters.

Show the answer

With the stored constants, $H(12)$ $= a + b\ln 12$ $\approx 45.293$ centimeters. Do not round $a$ and $b$ before this step.

Answer: 45.293 centimeters

6. Monthly sales are modeled by $S(t) = 30 + 12\ln t$ , in thousands of dollars, $t$ months after launch. The model's average rate of change on $[1, 10]$ is greater than on $[10, 19]$ . What does this mean in context?

Sales keep increasing, but each later month adds less than earlier months did
Sales start to decrease after month $10$ , so the model has turned downward
Sales increase faster after month $10$ than they did before month $10$
Sales reach their highest value at month $10$ and then stay at that level

Show the answer

Both rates are positive, so sales rise. The rate is smaller on the later interval, so the rate of change of $S$ is decreasing: growth slows down.

Correct: Sales keep increasing, but each later month adds less than earlier months did: Positive but decreasing rates of change.
Sales start to decrease after month $10$ , so the model has turned downward: The average rate on $[10, 19]$ is still positive, so $S$ is increasing.
Sales increase faster after month $10$ than they did before month $10$ : The later rate is smaller, so growth slows.
Sales reach their highest value at month $10$ and then stay at that level: $S$ is increasing for every $t > 0$ , so it has no maximum.

7. Reading speed is modeled by $y = 52 + 18\ln x$ words per minute after $x$ weeks. Since $18\ln 2 \approx 12.5$ , what does the model say?

Each doubling of the number of weeks adds about $12.5$ words per minute
Each week the speed rises by about $12.5$ words per minute, at a steady linear pace
The reading speed doubles once each $12.5$ weeks, as in exponential growth
After $2$ weeks of practice, the reading speed is about $12.5$ words per minute

Show the answer

$y(2x) - y(x)$ $= 18\ln 2$ $\approx 12.5$ : doubling the input adds the same amount to the output.

Correct: Each doubling of the number of weeks adds about $12.5$ words per minute: A fixed factor on the input adds a fixed amount to the output.
Each week the speed rises by about $12.5$ words per minute, at a steady linear pace: A constant increase per week would be linear.
The reading speed doubles once each $12.5$ weeks, as in exponential growth: Doubling outputs would be exponential.
After $2$ weeks of practice, the reading speed is about $12.5$ words per minute: $y(2)$ $= 52 + 18\ln 2$ $\approx 64.5$ .

Part 9 · Summary

Summary

Use a logarithmic model

y = a + b\ln x

when the input must multiply by a constant for the output to rise by a constant. Build it from two points or by logarithmic regression, keeping the constants stored. Predict outputs by substituting and find inputs by rewriting as a power. The model is concave down, so its rate of change decreases; it is supported only near the data and fails near

x = 0

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic