Exponential Modeling — AP® Precalculus Notes

When a quantity changes by the same factor over equal stretches of time, an exponential model $f(t) = ab^t$ describes it: a phone losing value, a medicine leaving the blood, a gym gaining members. This page builds such models three ways, rewrites them for other time units, and uses them to predict and to find when a value is reached. It is the content of Question 2 whenever the context is exponential.

Reading $a$ and $b$ in context

$a = f(0)$ is the initial value, the amount when $t = 0$ . $b$ is the growth factor per unit of $t$ : each unit, the amount is multiplied by $b$ . A factor $b = 1.035$ means a $3.5\%$ increase per unit; $b = 0.8$ means the amount keeps $80\%$ , a $20\%$ decrease. Always say what $t = 0$ is and what one unit of $t$ is.

Three ways to build the model

You know	Find $a$ and $b$ by
The initial value and the percent change	$a$ = initial value, $b = 1 + r$
Two points	The ratio of the outputs over the gap in inputs: $b^{\text{gap}} = \text{ratio}$
Many data points that fit only roughly	Exponential regression on the calculator

Worked example 1 (two points). A phone costs $$800$ new and is worth $$512$ after $2$ years. Find $V(t) = ab^t$ .

$a = V(0) = 800$ .

Over the $2$ -year gap the value is multiplied by $\frac{512}{800} = 0.64$ . That is two yearly factors: $b^2 = 0.64$ .

Take the square root (rational exponent $\frac{1}{2}$ ): $b = 0.64^{1/2} = 0.8$ . So $V(t) = 800(0.8)^t$ : the phone keeps $80\%$ of its value each year.

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. — **Figure 1.** $V(t) = 800(0.8)^t$ through $(0, 800)$ and $(2, 512)$ . LevlPrep original diagram.

If neither point is at $t = 0$ , find $b$ the same way, then work back to $a$ . Through $(1, 30)$ and $(3, 270)$ : $b^2 = 9$ , so $b = 3$ , and $a \cdot 3^1 = 30$ gives $a = 10$ .

With real data, use exponential regression: enter the lists, run the regression, and store $a$ and $b$ at full precision. For the medicine data on the practice page, the calculator gives $b = 0.79686...$ , a decrease of about $20.314\%$ per hour.

Other time units

Rational exponents convert the factor to another unit. If the yearly factor is $1.2$ , the monthly factor $m$ satisfies $m^{12} = 1.2$ , so $m = 1.2^{1/12}$ . The model can be written $P(t)$ $= 1200(1.2^{1/12})^{12t}$ , which is the same function as $1200(1.2)^t$ .

Do not split the percent: $20\%$ a year is not $\frac{20}{12}\%$ a month, because each month builds on the last. Likewise, $30\%$ a week is a daily factor of $1.3^{1/7} \approx 1.038$ , not $1 + \frac{0.3}{7}$ .

A half-life $h$ means the amount halves every $h$ units: $A(t) = A_0\left(\frac{1}{2}\right)^{t/h}$ . After $15$ years with a $5$ -year half-life, $80$ grams becomes $80\left(\frac{1}{2}\right)^3 = 10$ grams.

Predicting and solving

Worked example 2. For $V(t) = 800(0.8)^t$ , find $V(3.5)$ and the time when the phone is worth $$100$ .

Prediction: $V(3.5)$ $= 800(0.8)^{3.5}$ $= 366.357...$ , about $$366.36$ .

Solving: graph $y = 800(0.8)^t$ and $y = 100$ and find the intersection with the calculator: $t = 9.318...$ , so about $9.319$ years. (Store the value; do not round $b$ first.)

Where the model can be trusted

A model built from data is supported near that data. Between data points, predictions are usually good. Far past the last data point, nothing in the data shows the pattern continues: the phone may stop losing value at a scrap price, and the body may clear a medicine differently once little is left. Say this with the numbers: "the data stop at $t = 5$ , so a prediction at $t = 24$ is not supported by the data." In context, the domain may also have a hard edge, such as a value that cannot fall below a floor.