Exponential Growth and Decay worksheet with math exercises and examples.

Let's solve each problem step by step using the exponential growth and decay formula:

🔷 General Formula:

$$
y = a(1 + r)^t
$$
Where:
- $ y $: final amount after time $ t $
- $ a $: initial amount (starting value)
- $ r $: growth rate (as a decimal, e.g., 6% → 0.06)
- $ t $: time in years

For decay, $ r $ is negative.

---

✔ Problem 1: Investment Growth

> The original value of an investment is $8000, and the value increases by 5% each year. Write an exponential growth function to model this situation. Then, find the value of the investment after 20 years.

#### Step 1: Identify values
- $ a = 8000 $
- $ r = 5\% = 0.05 $
- $ t = 20 $

#### Step 2: Write the function
$$
y = 8000(1 + 0.05)^t = 8000(1.05)^t
$$

#### Step 3: Find value after 20 years
$$
y = 8000(1.05)^{20}
$$
Calculate:
$$
(1.05)^{20} \approx 2.6533
$$
$$
y \approx 8000 \times 2.6533 = 21,226.40
$$

✔ Answer:
Function: $ y = 8000(1.05)^t $
Value after 20 years: $21,226.40

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✔ Problem 2: College Tuition Increase

> The cost of tuition at a college is $12,000 and is increasing at a rate of 6% each year. Write an exponential growth function to model this situation. Then, find the tuition cost after 4 years.

#### Step 1: Values
- $ a = 12000 $
- $ r = 6\% = 0.06 $
- $ t = 4 $

#### Step 2: Function
$$
y = 12000(1 + 0.06)^t = 12000(1.06)^t
$$

#### Step 3: After 4 years
$$
y = 12000(1.06)^4
$$
$$
(1.06)^4 \approx 1.2625
$$
$$
y \approx 12000 \times 1.2625 = 15,150
$$

✔ Answer:
Function: $ y = 12000(1.06)^t $
Tuition after 4 years: $15,150

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✔ Problem 3: Student Enrollment Growth

> The number of student athletes at a local high school is 300 and is increasing at a rate of 8% per year. Write an exponential growth function to model this situation. Then, find the number of student athletes after 7 years.

#### Step 1: Values
- $ a = 300 $
- $ r = 8\% = 0.08 $
- $ t = 7 $

#### Step 2: Function
$$
y = 300(1 + 0.08)^t = 300(1.08)^t
$$

#### Step 3: After 7 years
$$
y = 300(1.08)^7
$$
$$
(1.08)^7 \approx 1.7138
$$
$$
y \approx 300 \times 1.7138 = 514.14 \Rightarrow \text{round to } 514
$$

✔ Answer:
Function: $ y = 300(1.08)^t $
Number after 7 years: 514 students

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✔ Problem 4: Car Depreciation (Decay)

> Anna buys a car for $19,000 and it depreciates at a rate of 7% per year. Write an exponential growth function to model this situation. Then, find the annual value after 7 years.

⚠️ Note: This is decay, so use $ r = -0.07 $

#### Step 1: Values
- $ a = 19000 $
- $ r = -7\% = -0.07 $
- $ t = 7 $

#### Step 2: Function
$$
y = 19000(1 - 0.07)^t = 19000(0.93)^t
$$

#### Step 3: After 7 years
$$
y = 19000(0.93)^7
$$
$$
(0.93)^7 \approx 0.6274
$$
$$
y \approx 19000 \times 0.6274 = 11,920.60
$$

✔ Answer:
Function: $ y = 19000(0.93)^t $
Value after 7 years: $11,920.60

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✔ Problem 5: Town Population Growth

> The population of a small town is 1000 and is increasing at a rate of 7% per year. Write an exponential growth function to model this situation. Then, find the population of the town after 10 years.

#### Step 1: Values
- $ a = 1000 $
- $ r = 7\% = 0.07 $
- $ t = 10 $

#### Step 2: Function
$$
y = 1000(1 + 0.07)^t = 1000(1.07)^t
$$

#### Step 3: After 10 years
$$
y = 1000(1.07)^{10}
$$
$$
(1.07)^{10} \approx 1.9672
$$
$$
y \approx 1000 \times 1.9672 = 1,967.2 \Rightarrow \text{round to } 1,967
$$

✔ Answer:
Function: $ y = 1000(1.07)^t $
Population after 10 years: 1,967 people

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✔ Problem 6: Bacteria Growth (from 1985 to 2000)

> In 1985, there were 385 coli plasmid bacteria in Hepatitis. The number of subcultures increased by 75% per year after 1985. How many subcultures were in Hepatitis in 2000?

#### Step 1: Time span
From 1985 to 2000 = $ 15 $ years

#### Step 2: Values
- $ a = 385 $
- $ r = 75\% = 0.75 $
- $ t = 15 $

#### Step 3: Function
$$
y = 385(1 + 0.75)^{15} = 385(1.75)^{15}
$$

#### Step 4: Calculate
First, compute $ (1.75)^{15} $

Use logarithms or calculator:
$$
(1.75)^{15} \approx 1,477,455.1
$$
Then:
$$
y \approx 385 \times 1,477,455.1 \approx 570,324,213.5
$$

That’s a huge number — likely due to extreme growth.

But let’s double-check if this makes sense.

Wait — is 75% per year realistic? Yes, for bacteria in ideal conditions.

So we proceed.

✔ Answer:
Function: $ y = 385(1.75)^t $
In 2000 ($ t = 15 $): ≈ 570,324,214 subcultures

(Note: This is a theoretical calculation; actual biology may differ.)

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📌 Summary of Answers:

| Problem | Function | Answer |
|--------|---------|--------|
| 1 | $ y = 8000(1.05)^t $ | $ 21,226.40 $ |
| 2 | $ y = 12000(1.06)^t $ | $ 15,150 $ |
| 3 | $ y = 300(1.08)^t $ | $ 514 $ |
| 4 | $ y = 19000(0.93)^t $ | $ 11,920.60 $ |
| 5 | $ y = 1000(1.07)^t $ | $ 1,967 $ |
| 6 | $ y = 385(1.75)^t $ | $ \approx 570,324,214 $ |

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