Topic 2.5A · CED: Exponential Function Context and Data Modeling

Exponential Context and Data Modeling

So far the models were handed to you: $N(t) = 256(1.5)^{t}$ , $C(t) = 160(0.87)^{t}$ . Real problems don't start with a formula. They start with a sentence (“the app grows 6% a week”, “caffeine has a half-life of about 5 hours”) or with a few measurements. This note is about turning that information into a model, and then reading the model back in the language of the situation.

11 MIN READ7 IDEAS34 PROBLEMS18 flashcards

Practice 34 problems Next: Chapter 2.5B Exponential Function Context and Data Modeling

Every exponential model needs the same two ingredients: a starting amount a, and a factor $b$ for some fixed length of time. Each section below is a different way those two ingredients show up. (2.5B covers models built from a whole data set.)

From a Percent Rate

CONCEPT

Growth or Decay by a Percent

If a quantity starts at a and grows by $r$ (as a decimal) each time unit, it is multiplied by $1 + r$ each unit. If it decays by $r$ each unit, only $1 - r$ of it remains, so it is multiplied by $1 - r$ .

Remember from 2.3: the base $b$ is what you HAVE after one step, not what you gain or lose.

f(t) = a\,(1 + r)^{t} \qquad\qquad f(t) = a\,(1 - r)^{t}

Worked example

Example 1. A study app has 3,200 users and grows 6% per week. A laptop bought for $1,450 loses 18% of its value each year. Write both models and evaluate $U(10)$ and $L(4)$ .

01
App: $a = 3200$ and $b = 1 + 0.06 = 1.06$ , so $U(w) = 3200(1.06)^{w}$ . After 10 weeks, $U(10) = 3200(1.06)^{10} \approx 5,730.71$ , so about 5,731 users.
02
Laptop: $a = 1450$ and $b = 1 - 0.18 = 0.82$ , so $L(t) = 1450(0.82)^{t}$ . After 4 years, $L(4)$ ≈ $655.58.
03
Sanity check: the app gained about 79% in 10 weeks ( $1.06^{10} \approx 1.79$ ), more than $10 \times 6$ % = 60%, because each week's 6% is taken of a bigger number.

COMMON MISTAKE

“6% growth means $b = 1.6$ ” and “18% decay means $b = 0.18$ .”

1.6 is 60% growth: with it the app would reach about 351,844 users in 10 weeks. And 0.18 would keep only 18% of the laptop's value: after one year it would be worth $261.00 instead of $1,189.00. Write the percent as a decimal first (6% = 0.06, 18% = 0.18), then add it to 1 or subtract it from 1.

From a Half-Life or a Doubling Time

Sometimes the rate isn't given per unit. Instead you're told how long it takes to halve or to double.

CONCEPT

Half-Life and Doubling Time

Half-life $h$ : the time for the amount to be cut in half. After $t$ units of time, t/h half-lives have passed, so the amount has been multiplied by $1/2$ a total of t/h times.

Doubling time $d$ : the time for the amount to double. After $t$ units, it has doubled t/d times.

f(t) = a \cdot \left(\tfrac{1}{2}\right)^{t/h} \qquad\qquad f(t) = a \cdot 2^{t/d}

REAL-LIFE EXAMPLE

Caffeine, Rewritten

Caffeine's half-life in the body is about 5 hours. Starting from 160 mg, $C(t) = 160 \cdot (1/2)^{t/5}$ : 80 mg after 5 hours, 40 after 10, 20 after 15.

Remember from 2.4: the power rule turns this into an hourly factor. $(1/2)^{1/5} \approx 0.8706$ , which is almost exactly the 0.87 from 2.3. The two models describe the same fading; one talks in half-lives, the other in hours.

Every 5 hours the amount is cut in half, whatever the starting amount at that moment.

C(t) = 160 \cdot \left(\tfrac{1}{2}\right)^{t/5} = 160 \cdot \left(\left(\tfrac{1}{2}\right)^{1/5}\right)^{t} \approx 160\,(0.8706)^{t}

Why t/h? The exponent counts how many half-lives have passed. For caffeine ( $h = 5$ ):

t (hours)	0	5	7.5	10	15
t/5 (half-lives)	0	1	1.5	2	3
C(t) (mg)	160	80	≈ 56.57	40	20

In between whole half-lives the formula still works: after 7.5 hours, 1.5 half-lives have passed, and $160(1/2)^{1.5} \approx 56.57$ mg.

Doubling works the same way. A mold colony covering 40 cm² that doubles every 3 days is $A(t) = 40 \cdot 2^{t/3}$ . After 9 days (3 doublings) $A(9) = 320$ cm², and after 10 days $A(10) \approx 403.17$ cm². The daily factor is $2^{1/3} \approx 1.2599$ , about 26% growth per day.

For Nova, $1.5^{1} < 2 < 1.5^{2}$ , so its doubling time is between 1 and 2 months (about 1.71). Finding it exactly needs logarithms, which you will use in 2.13.

Quick check

A 96 mg sample has a half-life of 4 hours. How much remains after 12 hours?

From Two Data Points

Given two points on $f(x) = a \cdot b^{x}$ , write one equation for each point, then DIVIDE them. Dividing cancels a and leaves a single power of $b$ .

Worked example

Example 2 (odd gap). A tracer in a lab sample measures 50 units on day 2 and 3.2 units on day 5. Model it as $f(t) = a \cdot b^{t}$ .

01
Two equations, then divide:
$a\,b^{2} = 50, \quad a\,b^{5} = 3.2 \quad\Rightarrow\quad \frac{a\,b^{5}}{a\,b^{2}} = b^{3} = \frac{3.2}{50} = 0.064$
02
$b^{3} = 0.064$ . A cube root has exactly one real answer: $b = 0.4$ . The tracer keeps 40% (loses 60%) each day.
03
Substitute back: $a \cdot 0.4^{2} = 50$ , so $a = 50 \div 0.16 = 312.5$ . Model: $f(t) = 312.5(0.4)^{t}$ . Check: $312.5(0.4)^{5} = 3.2$ ✓.

Worked example

Example 3 (even gap). A channel, Echo, had 1,800 subscribers at month 1 and 7,200 at month 3. Model it as $f(t) = a \cdot b^{t}$ .

01
Divide the two equations:
$\frac{a\,b^{3}}{a\,b^{1}} = b^{2} = \frac{7200}{1800} = 4 \quad\Rightarrow\quad b = 2 \;\text{ or }\; b = -2$
02
An even power hides the sign, so there are TWO algebraic answers. Now decide. An exponential function needs $b > 0$ (2.2, 2.3): with $b = -2$ , the value at $t = 1/2$ would need the square root of a negative number. The context agrees: $b = -2$ forces $a = 1800 \div (-2) = -900$ , and then month 2 would have $-900 \cdot (-2)^{2} = -3,600$ subscribers. So $b = 2$ .
03
$a \cdot 2 = 1800$ gives $a = 900$ . Model: $f(t) = 900(2)^{t}$ . Echo doubles every month. Check: $900 \cdot 2^{3} = 7,200$ ✓.

Both $b = 2$ and $b = -2$ pass through the two data points, but only $b = 2$ gives a real exponential function with sensible values.

COMMON MISTAKE

“ $b^{2} = 4$ , so $b = 2$ .” (Right answer, missing reason.)

Writing only the positive root gets the right model here, but for the wrong reason: it skips a step that sometimes matters. In 2.1 a sequence with $r = -3$ was perfectly valid. Always write both roots, $b = \pm 2$ , and then state why one is rejected: an exponential function's base must be positive (and here, subscriber counts cannot be negative).

“ $b^{2} = 4$ , so $b = 4 \div 2 = 2$ .”

This happens to give 2, but only by coincidence. Try $b^{4} = 16$ : dividing gives 4, while the correct 4th root is 2. Take the root that matches the exponent.

Worked example

Try it yourself. Each function is exponential, $f(x) = a \cdot b^{x}$ . (a) $f(1) = 3$ and $f(7) = 192$ . (b) $f(0) = 5$ and $f(4) = 80$ . Find $a$ and $b$ , showing the rejected root.

Answers: (a) $b^{6} = 192/3 = 64$ , so $b = 2$ or $b = -2$ ; reject −2 (the base must be positive). $a = 3 \div 2 = 1.5$ , so $f(x) = 1.5(2)^{x}$ . Check: $1.5 \cdot 2^{7} = 192$ ✓. (b) $b^{4} = 80/5 = 16$ , so $b = \pm 2$ ; reject −2. $a = f(0) = 5$ , so $f(x) = 5(2)^{x}$ , and $f(3) = 40$ .

Models With Base e

Many science models are written as $a \cdot e^{kt}$ . Remember from 2.4 that $e^{kt} = (e^{k})^{t}$ , so this is still $a \cdot b^{t}$ , with $b = e^{k}$ . If $k > 0$ it is growth; if $k < 0$ it is decay.

Worked example

Example 4. The views of a video $h$ hours after posting are modeled by $V(h) = 1200 e^{0.25h}$ . Find the hourly growth factor and percent, and $V(6)$ .

01
Rewrite with the power rule:
$V(h) = 1200\,e^{0.25h} = 1200\,\left(e^{0.25}\right)^{h} \approx 1200\,(1.2840)^{h}$
02
$b = e^{0.25} \approx 1.2840$ , so the views grow by about 28.4% per hour. $V(6) = 1200 e^{1.5} \approx 5,378.03$ views.
03
A negative $k$ means decay. For example, $500 e^{-0.2t} = 500(e^{-0.2})^{t} \approx 500(0.8187)^{t}$ : the factor is about 0.8187, so the amount drops about 18.1% per unit of $t$ .
04
Decay works the same way: $e^{-0.139t} = (e^{-0.139})^{t} \approx 0.8702^{t}$ , which is caffeine's hourly factor again.

COMMON MISTAKE

“ $k = 0.25$ , so the views grow 25% per hour.”

$k$ is not the percent. The factor is $e^{k} \approx 1.2840$ , so the rate is about 28.4% per hour. $k$ and the percent are close only when $k$ is small.

Quick check

A model is $M(t) = 30e^{0.12t}$ . What is the growth factor per unit of $t$ , to four decimals?

Reading a Model in Context

A model is only useful if you can say what its numbers mean, with units.

CONCEPT

Interpreting $a \cdot b^{t}$

a: the amount at $t = 0$ . For $V$ : 1,200 views when the video is posted.

$b$ : the factor per ONE unit of $t$ . For $V$ : each hour, the views are multiplied by about 1.284.

Average rate of change over [ $t_{1}$ , $t_{2}$ ]: ( $f(t_{2}) - f(t_{1})$ ) ÷ ( $t_{2} - t_{1}$ ), in output units per input unit.

\frac{V(6) - V(2)}{6 - 2} \approx \frac{5378.03 - 1978.47}{4} \approx 849.89

From hour 2 to hour 6, the video gained on average about 849.89 views per hour. Because the function is concave up, the rate is smaller than this near hour 2 and larger near hour 6, so this is an average, not a constant speed.

Practice

Worked example

P1. A town has 2,500 residents and grows 3.5% per year. Write a model and predict the population after 8 years.

Answer: $P(t) = 2500(1.035)^{t}$ . $P(8) \approx 3,292.02$ , so about 3,292 residents.

Worked example

P2. A medication has a half-life of 12 days. How much of an 800 mg dose remains after 30 days?

Answer: $M(t) = 800(1/2)^{t/12}$ . 30 days is 2.5 half-lives: $M(30) = 800(0.5)^{2.5} \approx 141.42$ mg.

Worked example

P3. $f$ is exponential with $f(2) = 36$ and $f(4) = 81$ . Find $f(x)$ and $f(6)$ .

Answer: $b^{2} = 81/36 = 9/4$ , so $b = 3/2$ or −3/2; reject −3/2 (base must be positive). $a = 36 \div (3/2)^{2} = 16$ , so $f(x) = 16(1.5)^{x}$ and $f(6) = 182.25$ .

Worked example

P4. A bacteria culture starts with 500 cells and doubles every 4 hours. Write a model, find the hourly growth factor, and find the population after 10 hours.

Answer: $B(t) = 500 \cdot 2^{t/4}$ . Hourly factor $2^{1/4} \approx 1.1892$ (about 18.9% per hour). $B(10) = 500 \cdot 2^{2.5} \approx 2,828.43$ cells.

Common slips

“6% growth means $b = 1.6$ ” and “18% decay means $b = 0.18$ .”

1.6 is 60% growth: with it the app would reach about 351,844 users in 10 weeks. And 0.18 would keep only 18% of the laptop's value: after one year it would be worth $261.00 instead of $1,189.00. Write the percent as a decimal first (6% = 0.06, 18% = 0.18), then add it to 1 or subtract it from 1.
“ $b^{2} = 4$ , so $b = 2$ .” (Right answer, missing reason.)

Writing only the positive root gets the right model here, but for the wrong reason: it skips a step that sometimes matters. In 2.1 a sequence with $r = -3$ was perfectly valid. Always write both roots, $b = \pm 2$ , and then state why one is rejected: an exponential function's base must be positive (and here, subscriber counts cannot be negative).

“ $b^{2} = 4$ , so $b = 4 \div 2 = 2$ .”

This happens to give 2, but only by coincidence. Try $b^{4} = 16$ : dividing gives 4, while the correct 4th root is 2. Take the root that matches the exponent.
“ $k = 0.25$ , so the views grow 25% per hour.”

$k$ is not the percent. The factor is $e^{k} \approx 1.2840$ , so the rate is about 28.4% per hour. $k$ and the percent are close only when $k$ is small.

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18 cards · Linear or exponential?, Exponential models

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Recap card

6 lines to re-read the night before.

01
Percent: growth by $r$ → $b = 1 + r$ ; decay by $r$ → $b = 1 - r$ . The base is what remains, not the change.
02
Half-life $h$ : $a(1/2)^{t/h}$ . Doubling time $d$ : $a \cdot 2^{t/d}$ . The exponent t/h counts how many halvings have happened.
03
Two points: write $a \cdot b^{x}$ at each, divide to cancel a, then take the root that matches the gap.
04
Even gap: write both roots ±, then reject the negative one because an exponential base must be positive (and state the reason).
05
$a \cdot e^{kt} = a(e^{k})^{t}$ , so the factor is $e^{k}$ , not $1 + k$ .
06
Next, in 2.5B: building a model from a whole data set with exponential regression, and data that level off at a value other than 0.

Practice 34 problems Next: Chapter 2.5B Exponential Function Context and Data Modeling

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Exponential Context and Data Modeling

From a Percent Rate

From a Half-Life or a Doubling Time

From Two Data Points

Models With Base e

Reading a Model in Context

Practice

Recap card

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