Exponential Growth and Decay worksheet with problems on calculating growth rates and applying exponential models.

Let’s solve each part of the worksheet step by step.

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Part 1: Given the growth/decay percentage, determine the multiplier.

The multiplier is found by converting the percent to a decimal and then:

- For growth: add it to 1 → multiplier = 1 + (percent as decimal)
- For decay: subtract it from 1 → multiplier = 1 - (percent as decimal)

a) 3% increase
→ 3% = 0.03
→ multiplier = 1 + 0.03 = 1.03

b) 52% decrease
→ 52% = 0.52
→ multiplier = 1 - 0.52 = 0.48

c) 200% increase
→ 200% = 2.00
→ multiplier = 1 + 2.00 = 3.00

d) 0.5% decrease
→ 0.5% = 0.005
→ multiplier = 1 - 0.005 = 0.995

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Part 2: State whether each function represents growth or decay.

We look at the base of the exponent (the number being raised to the power x). If it’s > 1 → growth. If it’s between 0 and 1 → decay.

a) f(x) = 10 · 3^x
→ Base = 3 → greater than 1 → Growth

b) f(x) = 60 · (0.2)^x
→ Base = 0.2 → less than 1 → Decay

c) f(x) = 8 · e^(0.7x)
→ This is exponential with base e ≈ 2.718, and exponent is positive (0.7x) → so it grows → Growth

d) f(x) = 100 · (0.99)^x
→ Base = 0.99 → less than 1 → Decay

e) f(x) = 2 · e^(-0.5x)
→ Exponent is negative → this means it decays → Decay

f) f(x) = 5 · (1.01)^x
→ Base = 1.01 → greater than 1 → Growth

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Part 3: Identify necessary information and write an equation for each scenario.

a) Bacteria Culture Problem

Given:
- Starts with 50 bacteria → initial amount = 50
- Doubles every hour → that means after 1 hour: 50 × 2, after 2 hours: 50 × 2², etc.
- So, growth factor per hour = 2
- Time in hours = x

Equation:
f(x) = 50 · 2^x

Questions:

i. Is this growth or decay?
→ Since we’re multiplying by 2 each time → Growth

ii. What is the starting value?
→ 50

iii. What is the rate of change?
→ It doubles every hour → so the rate is 100% per hour (because doubling = increasing by 100%)

iv. Write the explicit equation.
→ Already did: f(x) = 50 · 2^x

v. How many bacteria are there after 6 hours?
→ Plug in x = 6:
f(6) = 50 · 2⁶ = 50 · 64 = 3200

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b) Car Depreciation Problem

Given:
- Initial value = $20,000
- Loses 20% of its value each year → so it keeps 80% of its value each year
- That means multiplier = 1 - 0.20 = 0.80
- Time in years = x

Equation:
f(x) = 20000 · (0.80)^x

Questions:

i. What is your multiplier?
→ 0.80

ii. Is this growth or decay?
→ Losing value → Decay

iii. After how many years will the car be worth $10,000?
→ We need to solve:
20000 · (0.80)^x = 10000
Divide both sides by 20000:
(0.80)^x = 0.5
Take log of both sides:
log((0.80)^x) = log(0.5)
x · log(0.80) = log(0.5)
x = log(0.5) / log(0.80)
Using calculator:
log(0.5) ≈ -0.3010
log(0.80) ≈ -0.0969
x ≈ (-0.3010) / (-0.0969) ≈ 3.106

So about 3.1 years — but since the question likely expects whole years, we can say after 3 years it’s still above $10k, after 4 years below. But let’s check:

After 3 years: 20000 · (0.8)^3 = 20000 · 0.512 = $10,240
After 4 years: 20000 · (0.8)^4 = 20000 · 0.4096 = $8,192

So it crosses $10,000 between year 3 and 4. The exact answer is approximately 3.1 years, but if they want the first full year when it’s under $10k, it’s year 4. However, since the question says “after how many years”, and doesn’t specify rounding, we’ll give the precise value.

But note: In many school contexts, they might expect you to use logs and report the decimal. Let’s go with approximately 3.1 years.

iv. Write the explicit equation.
→ f(x) = 20000 · (0.80)^x

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Final Answer:

Part 1:
a) 1.03
b) 0.48
c) 3.00
d) 0.995

Part 2:
a) Growth
b) Decay
c) Growth
d) Decay
e) Decay
f) Growth

Part 3a:
i. Growth
ii. 50
iii. 100% per hour
iv. f(x) = 50 · 2^x
v. 3200

Part 3b:
i. 0.80
ii. Decay
iii. Approximately 3.1 years
iv. f(x) = 20000 · (0.80)^x