Exponential Modeling — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

A new phone costs $800. Two years later it sells for $512. Assume it keeps losing the same fraction of its value each year, and two numbers become a model that predicts its value at any time, and how long until it is worth $100.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A quantity decreases by $15\%$ each year. What is the decay factor?

$0.85$
$0.15$
$1.15$
$-0.15$

Show the answer

$1 - 0.15 = 0.85$ .

Correct: $0.85$ :
$0.15$ :
$1.15$ :
$-0.15$ :

2. Rewrite $1.2^{t}$ using a monthly factor, with $t$ in years.

$(1.2^{1/12})^{12t}$
$1.2^{t/12}$
$(\frac{1.2}{12})^{12t}$
$1.2^{12t}$

Show the answer

$(1.2^{1/12})^{12t} = 1.2^{t}$ by the power rule.

Correct: $(1.2^{1/12})^{12t}$ :
$1.2^{t/12}$ :
$(\frac{1.2}{12})^{12t}$ :
$1.2^{12t}$ :

Part 4 · See it

See it first

A phone's value V in dollars against years t: the points (0, 800) and (2, 512) on the decay curve V(t) = 800(0.8)^t, which falls toward the dashed asymptote V = 0. — Two data points, $(0, 800)$ and $(2, 512)$ , give $V(t) = 800(0.8)^t$ : the phone keeps $80\%$ of its value each year. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Know the initial value and the ratio, or two points, or many data pointsfind $a$ and $b$ directly, by the ratio over the gap, or by exponential regression
Read the base $b$ as a growth factorpercent change $(b - 1) \cdot 100$ per unit; a $k$ -unit factor $b^k$ , a $\frac{1}{k}$ -unit factor $b^{1/k}$
Use the stored modelpredict an output, or find the input for a given output with the calculator, inside the domain the context allows

Part 6 · Key ideas

Key ideas

Two points $(0, a)$ and $(k, y)$ : $b^k = \frac{y}{a}$ , so $b = \left(\frac{y}{a}\right)^{1/k}$ .
Exponential regression on the calculator fits $y = ab^x$ to all the data; store $a$ and $b$ .
Other time units: a yearly factor $b$ is a monthly factor $b^{1/12}$ ; a half-life $h$ gives $\left(\frac{1}{2}\right)^{t/h}$ .

Part 7 · Misconception

A common mistake

The wrong idea: A $30\%$ weekly growth is $\frac{30\%}{7}$ per day, a daily factor of $1 + \frac{0.3}{7}$ .

What actually happens: Growth compounds: the daily factor $d$ satisfies $d^7 = 1.3$ , so $d = 1.3^{1/7} \approx 1.038$ .

Part 8 · Check yourself

Check yourself

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

Table

Members of a climbing gym

Number of members P of a climbing gym t years after it opened
$t$ (years)	0	2
$P$ (members)	1200	1728

1. An exponential model $P(t) = ab^t$ fits the two data points. Write $P(t)$ .

Type an expression for $P(t) =$ .

Show the answer

$a = P(0) = 1200$ . Over $2$ years the ratio is $\frac{1728}{1200} = 1.44$ , so $b^2 = 1.44$ and $b = 1.2$ . $P(t) = 1200(1.2)^t$ .

Answer: $P(t) =$ $1200(1.2)^{t}$

2. By what percent does the membership grow each year, according to the model?

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

Show the answer

$b = 1.2 = 1 + 0.20$ , so the membership grows $20\%$ per year.

Answer: 20 percent

3. Which expression gives the monthly growth factor of the membership?

$1.2^{1/12}$
$\frac{1.2}{12}$
$1.2^{12}$
$1 + \frac{0.2}{12}$

Show the answer

A year is $12$ months, so $(\text{monthly factor})^{12} = 1.2$ and the monthly factor is $1.2^{1/12}$ . Then $P(t)$ $= 1200(1.2^{1/12})^{12t}$ with $t$ in years.

Correct: $1.2^{1/12}$ : Twelve equal monthly factors multiply to $1.2$ .
$\frac{1.2}{12}$ : Dividing the factor by $12$ gives $0.1$ , a loss.
$1.2^{12}$ : This is the factor over $12$ years.
$1 + \frac{0.2}{12}$ : Splitting the percent evenly ignores that each month builds on the last; $(1 + \frac{0.2}{12})^{12} \ne 1.2$ .

4. An exponential function passes through $(1, 30)$ and $(3, 270)$ . Write $f(x) = ab^x$ .

Type an expression for $f(x) =$ .

Show the answer

From $x = 1$ to $x = 3$ the output is multiplied by $\frac{270}{30} = 9$ over $2$ steps, so $b^2 = 9$ and $b = 3$ . Then $a \cdot 3 = 30$ gives $a = 10$ .

Answer: $f(x) =$ $10(3)^{x}$

5. A sample with a half-life of $5$ years has $80$ grams now. How many grams remain after $15$ years?

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

Show the answer

$15$ years is $3$ half-lives: $80 \cdot \left(\frac{1}{2}\right)^3 = 10$ grams.

Answer: 10 grams

6. The model $S(t) = 4.2(1.035)^t$ gives a city's average home price, in hundreds of thousands of dollars, $t$ years after 2020. What does $1.035$ mean?

The price increases by $$1.035$ each year
The price in 2020 was $1.035$ hundred thousand dollars
Each year the price rises $3.5\%$ over the year before
The price increases by $103.5\%$ each year

Show the answer

$1.035$ is the growth factor: each year the output is multiplied by $1.035 = 1 + 0.035$ , a $3.5\%$ increase.

The price increases by $$1.035$ each year: The factor multiplies; it is not an added amount.
The price in 2020 was $1.035$ hundred thousand dollars: The 2020 price is $S(0) = 4.2$ , about $$420{,}000$ .
Correct: Each year the price rises $3.5\%$ over the year before: Growth factor $1 + 0.035$ .
The price increases by $103.5\%$ each year: A $103.5\%$ increase would more than double the price.

Part 9 · Summary

Summary

Build

ab^t

from an initial value and a ratio, from two points, or by exponential regression. The base is the growth factor; rewrite it for other time units with rational exponents. Predict and solve with the stored model, and only where the context and data allow.

Part 10 · Up next

What comes next

Part 11 · Connections