Putting It Together: Growth and Decay Problem Solving

1) Choose the Model Type from Context (and Justify It)

Most growth/decay problems start with a context clue that tells you whether to use a discrete factor model (repeated multiplication at fixed intervals) or a continuous-like model (smooth change over time). Your first job is to translate the story into the right “time step.”

Discrete factor (compounded per interval)

• Signals: “per year,” “each month,” “every 5 minutes,” “compounded monthly,” “doubles every 3 days,” “decreases by 12% each week.”
• Model form: A(t)=A_0\,b^t where t counts the number of intervals (years, months, weeks, etc.).
• Interpretation: b is the multiplication factor each interval (e.g., 1.08 means +8% per interval; 0.93 means −7% per interval).

Continuous-like change (constant percent rate per unit time)

• Signals: “continuously,” “constant relative rate,” “instantaneous rate,” “half-life,” “decays continuously,” “grows at 5% per year (continuously).”
• Model form: A(t)=A_0 e^{kt} where t is time in chosen units and k is the continuous rate per unit time.
• Interpretation: If k>0 it grows; if k<0 it decays. The percent rate is approximately 100k% per unit time for small |k|, but the exact relationship to a discrete annual factor is b=e^k (for 1-year steps).

Quick decision checklist

• Does the quantity update at regular checkpoints? Use A_0 b^t.
• Does the context imply smooth change or give half-life/continuous rate? Use A_0 e^{kt}.
• Are you given a percent change “per period” with no mention of continuous? Default to discrete unless stated otherwise.

2) Solve for the Unknown (Initial Value, Factor/Rate, or Time)

Once the model type is chosen, identify what is unknown: initial value (A_0), factor/rate (b or k), or time (t). Keep units attached to every number.

A) Solve for the initial value A_0

Example (discrete): A population is 18,500 after 6 years, growing by 4% per year. Find the initial population.

Model: A(t)=A_0(1.04)^t. Plug in t=6, A(6)=18500:

18500 = A0(1.04)^6  =>  A0 = 18500/(1.04)^6

Compute and round appropriately (population is a count): A_0\approx 18500/1.2653\approx 14620. So the initial population was about 14,620 people.

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Example (continuous-like): A chemical sample has mass 12.0 g after 3.5 hours and decays with k=-0.18 per hour. Find the initial mass.

12.0 = A0 e^{-0.18(3.5)}  =>  A0 = 12.0 / e^{-0.63} = 12.0 e^{0.63}

A_0\approx 12.0(1.878)\approx 22.5. Initial mass ≈ 22.5 g.

B) Solve for the factor b or rate k

Example (find discrete factor): A device’s value drops from $900 to $540 in 4 years with a constant yearly factor. Find the yearly factor and percent decrease.

540 = 900 b^4  =>  b^4 = 0.6  =>  b = 0.6^{1/4}

b\approx 0.6^{0.25}\approx 0.880. That means about a 12.0% decrease per year (since 1-0.880=0.120).

Example (find continuous rate): A bacteria culture grows from 2,000 to 9,000 in 5 hours with continuous-like growth. Find k.

9000 = 2000 e^{5k}  =>  4.5 = e^{5k}  =>  ln(4.5) = 5k  =>  k = ln(4.5)/5

k\approx 1.5041/5\approx 0.3008 per hour. Interpreting: about 0.3008 hr⁻¹ continuous growth rate.

C) Solve for time t (log step)

Time is the unknown most likely to require a logarithm step. The workflow is: isolate the exponential part, then take a log.

Example (discrete time): An investment grows 6% per year. How long to double?

2 = (1.06)^t  =>  ln(2) = t ln(1.06)  =>  t = ln(2)/ln(1.06)

t\approx 0.6931/0.05827\approx 11.9. Interpretation: about 12 years to double (rounding to whole years makes sense for annual compounding).

Example (continuous-like time): A medication amount follows A(t)=A_0 e^{-0.22t}. How long until it reaches 25% of the initial amount?

0.25A0 = A0 e^{-0.22t}  =>  0.25 = e^{-0.22t}

ln(0.25) = -0.22t  =>  t = ln(0.25)/(-0.22)

t\approx (-1.3863)/(-0.22)\approx 6.30. About 6.3 hours.

3) Interpret Solutions: Units, Rounding, and Reasonableness Checks

Solving the equation is only half the job. Your answer should make sense in the original context.

Units and meaning

• State units explicitly: dollars, grams, people, years, hours.
• Interpret parameters: “factor 0.88 per year” means multiply by 0.88 each year; “k=-0.22 per hour” means the amount shrinks continuously at that relative rate.
• Be clear about what t counts: number of months vs years is not interchangeable.

Rounding decisions (context-driven)

• Counts (people, items): round to whole numbers, but consider whether rounding up/down changes the meaning (e.g., minimum time needed).
• Money: typically to cents, unless the context is large-scale estimates.
• Time to reach a threshold: if the question asks “how long until at least/at most,” rounding direction matters. If decay must drop below a level, you may need to round up to ensure the condition is met.

Reasonableness checks

• Growth vs decay sign check: If the quantity is supposed to decrease, your factor should be <1 (or k<0).
• Back-substitution check: Plug your answer back into the model to see if it reproduces the given value (approximately, if rounded).
• Scale check: If something “doubles” with 6% yearly growth, expecting around 12 years is plausible because 6% is modest; if you got 2 years, that’s a red flag.

4) Compare Two Exponential Models: Which Grows Faster and Why?

Comparing exponentials is about comparing their multiplication behavior. You can compare by (a) rewriting to a common base, (b) comparing growth rates, or (c) solving for the crossover time using logs.

A) Same time unit, same structure: compare bases or rates

If A(t)=A_0 b^t and C(t)=C_0 d^t with the same t unit, then for large t, the one with larger base (b vs d) eventually grows faster, but the initial values (A_0, C_0) can delay that.

B) Find the crossover time (algebra + log thinking)

Example: Plan A: A(t)=500(1.08)^t. Plan B: B(t)=900(1.05)^t. When does A exceed B?

Set them equal to find the crossover:

500(1.08)^t = 900(1.05)^t

(1.08/1.05)^t = 900/500 = 1.8

t ln(1.08/1.05) = ln(1.8)  =>  t = ln(1.8)/ln(1.08/1.05)

Compute: 1.08/1.05\approx 1.028571. Then t\approx 0.5878/0.02817\approx 20.9. Interpretation: after about 21 years, Plan A overtakes Plan B. Before that, B is larger due to the higher initial amount.

C) Comparing discrete vs continuous-like models

Sometimes one model is written with b^t and another with e^{kt}. You can still compare by converting one form to the other using b=e^k (same time unit). For example, e^{0.07t} corresponds to a discrete factor of e^{0.07}\approx 1.0725 per unit time, i.e., about 7.25% per unit time in a discrete-step sense.

5) Culminating Multi-Step Scenarios (Model → Solve → Interpret)

Scenario 1: Depreciation, then a threshold time

Context: A camera costs $1,600 new and loses 18% of its value each year. (a) Write a model for its value after t years. (b) When will it be worth $500 or less?

Step 1: Choose model type. “Each year” suggests discrete factor.

Step 2: Build model. Factor per year: b=1-0.18=0.82. So V(t)=1600(0.82)^t.

Step 3: Solve for time.

500 = 1600(0.82)^t  =>  0.3125 = (0.82)^t

t = ln(0.3125)/ln(0.82)

t\approx (-1.1632)/(-0.19845)\approx 5.86 years.

Step 4: Interpret with rounding direction. The question asks “$500 or less,” so you need the first whole year when the value is at most $500. Since t\approx 5.86, after 5 years it’s still above $500, and after 6 years it should be below. Report: about 6 years.

Scenario 2: Recover a starting amount from two measurements (continuous-like)

Context: A pollutant concentration decays continuously. After 2 days it is 48 mg/L, and after 5 days it is 30 mg/L. (a) Find a model C(t)=C_0 e^{kt} with t in days. (b) Find the initial concentration C_0. (c) Predict when it will reach 10 mg/L.

Step 1: Choose model type. “Decays continuously” suggests e^{kt}.

Step 2: Use the two data points to solve for k.

48 = C0 e^{2k}

30 = C0 e^{5k}

Divide the second by the first to eliminate C_0:

30/48 = e^{5k}/e^{2k} = e^{3k}

0.625 = e^{3k}  =>  ln(0.625)=3k  =>  k = ln(0.625)/3

k\approx (-0.4700)/3\approx -0.1567 per day.

Step 3: Solve for C_0. Use 48=C_0 e^{2k}:

C0 = 48 / e^{2k} = 48 e^{-2k}

With k\approx -0.1567, -2k\approx 0.3134, so C_0\approx 48e^{0.3134}\approx 65.7 mg/L.

Step 4: Predict time to 10 mg/L.

10 = 65.7 e^{-0.1567 t}  =>  10/65.7 = e^{-0.1567 t}

t = ln(10/65.7)/(-0.1567)

t\approx ln(0.1522)/(-0.1567)\approx (-1.882)/(-0.1567)\approx 12.0 days. Interpretation: about 12 days after the start, the concentration reaches 10 mg/L.

Scenario 3: Compare two savings plans and justify the better choice by a target date

Context: You want at least $20,000 in 10 years.

• Plan A: Start with $8,000 and earn 6.2% per year compounded annually.
• Plan B: Start with $10,000 and earn 5.4% per year compounded annually.

Step 1: Choose model type. “Compounded annually” is discrete.

Step 2: Write both models.

A(t)=8000(1.062)^t, B(t)=10000(1.054)^t.

Step 3: Evaluate at the target time t=10.

A(10)=8000(1.062)^{10}  ≈ 8000(1.826) ≈ 14608

B(10)=10000(1.054)^{10} ≈ 10000(1.692) ≈ 16920

Step 4: Interpret and reasonableness check. Neither plan reaches $20,000 by year 10. Plan B is larger at year 10 because it starts higher and the rate difference is not enough to catch up within 10 years. If the question were “which grows faster,” Plan A has the larger factor (1.062 vs 1.054) and will eventually overtake; but for the 10-year goal, Plan B yields more.

Scenario 4: Multi-step with a change of units (monthly vs yearly)

Context: A subscription base grows by 3% each month. It currently has 12,000 subscribers. (a) Model the number after m months. (b) How many months until it reaches 20,000? (c) Estimate how many years that is, and state your rounding choice.

Step 1: Choose model type. “Each month” indicates discrete monthly factor.

Step 2: Model. S(m)=12000(1.03)^m.

Step 3: Solve for months.

20000 = 12000(1.03)^m  =>  5/3 = (1.03)^m

m = ln(5/3)/ln(1.03)

m\approx 0.5108/0.02956\approx 17.3 months.

Step 4: Convert to years and interpret. 17.3 months is 17.3/12\approx 1.44 years. If the question is “until it reaches 20,000,” you might report 18 months (round up) because at 17 months it may still be below 20,000.

Scenario 5: A crossover comparison plus a short interpretation statement

Context: Two online creators start with different follower counts and grow at different rates.

• Creator X: 4,000 followers, grows 9% per month.
• Creator Y: 7,500 followers, grows 4% per month.

(a) Write models. (b) When will X surpass Y? (c) Write a one-sentence interpretation tied to the context.

Step 1: Model type. Monthly percent growth → discrete monthly factor.

Step 2: Models. X(m)=4000(1.09)^m, Y(m)=7500(1.04)^m.

Step 3: Solve crossover time.

4000(1.09)^m = 7500(1.04)^m

(1.09/1.04)^m = 7500/4000 = 1.875

m = ln(1.875)/ln(1.09/1.04)

1.09/1.04\approx 1.0480769, so m\approx 0.6286/0.04696\approx 13.4 months.

Step 4: Interpretation statement. Creator X is expected to surpass Creator Y after about 14 months because X’s higher monthly growth factor eventually outweighs Y’s larger starting follower count.

Mini reference table: what to isolate before taking logs