Building Exponential Models for Real Situations

1) Start from y = a·b^t and interpret the parameters

An exponential model for repeated (multiplicative) change over equal time steps is often written as y = a·b^t, where t counts the number of time steps since the start.

• a = initial value: the value at t = 0. It carries the same units as y (dollars, people, grams, °C, etc.).
• b = growth/decay factor per time step: how many times the quantity is multiplied each step. It is unitless, but it is tied to the chosen time step (per year, per month, per day…).
• t = number of time steps: must match the time step used by b. If b is “per month,” then t must be in months.

What values of b mean

• Growth: b > 1 (each step multiplies by more than 1).
• Decay: 0 < b < 1 (each step multiplies by a fraction).
• No change: b = 1 (constant).

Checkpoint: Always state the time step in words: “b = 1.03 per month” is different from “b = 1.03 per year.”

2) Create models from context statements and from data pairs

A) Building a model from a context statement

Many real situations tell you two things: the starting amount and the repeated percent change per time step. Translate directly into a and b.

Template

• Identify the initial value: “starts at …” → a.
• Identify the time step: per day/month/year/etc.
• Convert the percent change to a factor b (see Section 3).
• Write y(t) = a·b^t and label units.

Example: weekly decay

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A medication amount starts at 80 mg and decreases by 12% each week.

• a = 80 mg
• Decay factor per week: b = 1 − 0.12 = 0.88
• Model (with t in weeks): A(t) = 80·(0.88)^t mg

B) Building a model from two data pairs

If you know the value at two different times, you can solve for a and b. The key idea is that the ratio of values over a time gap equals b raised to that gap.

Suppose you know (t_1, y_1) and (t_2, y_2) and assume y = a·b^t.

• Divide the equations: y_2 / y_1 = (a·b^{t_2})/(a·b^{t_1}) = b^{t_2 - t_1}
• Solve for b: b = (y_2 / y_1)^{1/(t_2 - t_1)}
• Then solve for a using either point: a = y_1 / b^{t_1}

Step-by-step example (growth)

A town has 12,000 people at year 0 and 15,000 people at year 5. Assume exponential growth with yearly steps.

• t_1 = 0, y_1 = 12000; t_2 = 5, y_2 = 15000
• b = (15000/12000)^{1/5} = (1.25)^{1/5}
• Approximate: (1.25)^{1/5} ≈ 1.0456 (about 4.56% per year)
• Since t_1 = 0, a = 12000
• Model: P(t) = 12000·(1.0456)^t people, with t in years

Note on units: If the second data point were at 60 months instead of 5 years, you must keep the time step consistent (either convert 60 months to 5 years, or build a monthly model).

C) Building a model from a table of data (using ratios)

If data are measured at equal time steps, check whether the ratio y_{t+1}/y_t is roughly constant. If it is, that ratio is a good estimate for b.

Here the ratio is consistently 1.05, so a model is y = 500·(1.05)^t with t in months.

3) Compute percent growth/decay rates and convert between rate and factor

Real situations often describe change as a percent rate per time step. The exponential model uses a factor b. Convert between them carefully.

From percent rate r to factor b

• Growth by r (e.g., 7% growth): b = 1 + r
• Decay by r (e.g., 7% decay): b = 1 − r

Use r as a decimal: 7% → r = 0.07.

Examples

• 3% growth per year → b = 1.03
• 18% decay per month → b = 0.82

From factor b to percent rate r

• If b > 1, growth rate: r = b − 1
• If 0 < b < 1, decay rate: r = 1 − b

Example

If a quantity is multiplied by b = 0.94 each day, then the daily decay rate is r = 1 − 0.94 = 0.06 → 6% decay per day.

Time-step changes: do not “divide the percent”

If you have a factor for one time step and want a different time step, you must use roots/powers, not simple division of the percent.

Example: convert annual factor to monthly factor

If growth is 12% per year, then annual factor is b_year = 1.12. A consistent monthly factor b_month satisfies (b_month)^{12} = 1.12, so b_month = (1.12)^{1/12}.

This is why “1% per month” is not automatically the same as “12% per year.”

4) Use doubling/halving language as intuitive checkpoints

Doubling and halving provide quick reality checks for exponential models.

Doubling check

If a quantity doubles every D time steps, then b^D = 2, so b = 2^{1/D}.

Example

A bacteria culture doubles every 3 hours. Then b = 2^{1/3} ≈ 1.2599 per hour, and a model with t in hours is N(t) = N_0·(2^{1/3})^t.

Halving check

If a quantity halves every H time steps, then b^H = 1/2, so b = (1/2)^{1/H}.

Example

A chemical amount halves every 10 days. Then b = (1/2)^{1/10} ≈ 0.9330 per day, and A(t) = A_0·(0.9330)^t with t in days.

Checkpoint questions

• If your model predicts doubling but the context is “depreciation,” something is wrong (likely b should be less than 1).
• If the context says “halves every 5 years,” then after 10 years the factor should be about 1/4. Your model should reflect that: b^{10} ≈ 0.25.

5) Mini-case exercises (with units and time-step consistency)

Case 1: Population growth from a statement

Situation A city has 48,000 residents now and grows by 2.4% per year.

Build the model

• Initial value: a = 48000 people
• Growth factor per year: b = 1 + 0.024 = 1.024
• Model: P(t) = 48000·(1.024)^t, where t is in years

Use it Estimate population after 8 years:

P(8) = 48000·(1.024)^8 ≈ 48000·1.208 ≈ 57,984 people

Unit check The output is in people; the exponent uses years, matching “per year.”

Case 2: Savings with periodic growth (no deposits)

Situation You deposit $2,500 into an account that grows 0.6% per month.

Build the model

• a = 2500 dollars
• b = 1 + 0.006 = 1.006 per month
• S(t) = 2500·(1.006)^t, t in months

Use it Value after 3 years (36 months):

S(36) = 2500·(1.006)^{36} ≈ 2500·1.240 ≈ $3,100

Consistency checkpoint Do not plug t = 3 unless your factor is “per year.” Here it is “per month,” so use 36.

Case 3: Depreciation (value decreases by a percent each year)

Situation A laptop costs $1,400 new and loses 20% of its value each year.

Build the model

• a = 1400 dollars
• Decay factor per year: b = 1 − 0.20 = 0.80
• V(t) = 1400·(0.80)^t, t in years

Use it Value after 4 years:

V(4) = 1400·(0.80)^4 = 1400·0.4096 ≈ $573.44

Interpretation “20% per year” means multiply by 0.80 each year, not subtract $280 each year.

Case 4: Cooling-type decay without calculus (repeated fraction of the gap)

Some “cooling-like” situations are described as: “each time step, the object closes a fixed fraction of the gap to the surrounding temperature.” This produces an exponential model for the difference from ambient temperature.

Situation A drink is in a 20°C room. At the start it is 90°C. Each 5 minutes, it loses 15% of the difference between its temperature and room temperature.

Step 1: Define the difference variable

• Ambient (room) temperature: T_room = 20 °C
• Difference from room: D(t) = T(t) − 20 (in °C)

Step 2: Model the difference exponentially

“Loses 15% of the difference each step” means 85% of the difference remains each step:

• D(0) = 90 − 20 = 70 °C
• b = 1 − 0.15 = 0.85 per 5 minutes
• D(t) = 70·(0.85)^t, where t counts 5-minute intervals

Step 3: Convert back to temperature

T(t) = 20 + D(t) = 20 + 70·(0.85)^t

Use it After 20 minutes, that is t = 4 intervals:

T(4) = 20 + 70·(0.85)^4 ≈ 20 + 70·0.522 ≈ 56.5°C

Consistency checkpoint The factor 0.85 is per 5 minutes. If you want a per-minute factor, you would use 0.85^{1/5} per minute, and then t would be in minutes.

Case 5: Build a model from two data points (depreciation-style)

Situation A machine is worth $9,000 now (year 0) and $6,300 after 3 years. Assume exponential decay with yearly steps.

Find b

b = (6300/9000)^{1/3} = (0.7)^{1/3} ≈ 0.8879

Write the model

• a = 9000 dollars
• V(t) = 9000·(0.8879)^t, t in years

Interpret the rate

Yearly decay rate is r = 1 − b ≈ 1 − 0.8879 = 0.1121 → about 11.21% per year.

Practice set (mini-exercises)

• Population: A wildlife reserve has 2,300 deer and grows 5% per year. Write N(t) and estimate N(6). State units.
• Savings: $800 grows 1.2% per quarter. Write a model with t in quarters. What is the value after 5 years?
• Depreciation: A car worth $24,000 loses 14% per year. Write V(t). When will it drop below $10,000? (Solve by testing integer years.)
• Cooling-type: Room is 22°C. Soup starts at 78°C. Every 10 minutes it keeps 75% of its temperature difference from room. Write T(t). Find T(3) (after 30 minutes).
• Two-point model: A quantity is 500 at t=2 days and 350 at t=7 days. Build y=a·b^t with t in days (find b then a).