Exponential Growth And Decay Worksheet Outlet | www ... - Free Printable

Let’s solve each problem step by step.

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Problem 1: World Population Growth

Given:
- Population in 2000 = 6.08 billion
- Annual growth rate = 1.26% = 0.0126

a. Growth factor
The growth factor is 1 + growth rate (as a decimal).
So, growth factor = 1 + 0.0126 = 1.0126

b. Function to model population
We use the exponential growth formula:
P(x) = initial amount × (growth factor)^x
Where x = years after 2000
→ P(x) = 6.08 × (1.0126)^x (population in billions)

c. Population in 2010
2010 is 10 years after 2000 → x = 10
P(10) = 6.08 × (1.0126)^10

First, calculate (1.0126)^10:
Using calculator: 1.0126^10 ≈ 1.1337 (rounded to 4 decimals)
Then: 6.08 × 1.1337 ≈ 6.893 (billion)

✔ So, world population in 2010 ≈ 6.89 billion

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Problem 2: Computer Depreciation

Given:
- Initial value = $6500
- Depreciation rate = 14.3% per year → decay rate = 0.143

a. Function for value over time
Exponential decay: V(t) = initial × (1 - decay rate)^t
V(t) = 6500 × (1 - 0.143)^t = 6500 × (0.857)^t

b. Value after 3 years
t = 3
V(3) = 6500 × (0.857)^3

Calculate (0.857)^3:
0.857 × 0.857 = 0.734449
0.734449 × 0.857 ≈ 0.6294 (rounded)
Then: 6500 × 0.6294 ≈ 4091.10

✔ Value after 3 years ≈ $4,091.10

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Problem 3: Animal Population Decrease

Given:
- Initial count = 80 animals
- Decrease rate = 3.5% per year → decay rate = 0.035

a. Function for population change
P(t) = 80 × (1 - 0.035)^t = 80 × (0.965)^t
Where t = years since you started counting

b. Graph and estimate when population drops below 15

We need to find smallest t such that:
80 × (0.965)^t < 15

Divide both sides by 80:
(0.965)^t < 15/80 = 0.1875

Take log of both sides:
log((0.965)^t) < log(0.1875)
t × log(0.965) < log(0.1875)

Note: log(0.965) is negative, so inequality flips when dividing:

t > log(0.1875) / log(0.965)

Calculate:

log(0.1875) ≈ -0.727 (using base 10 or natural — same ratio)
log(0.965) ≈ -0.0153 (approx)

So: t > (-0.727) / (-0.0153) ≈ 47.5

✔ So, it takes about 48 years for population to drop below 15.

*(You can also test values: at t=47, P≈15.1; at t=48, P≈14.6 — confirms answer)*

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Problem 4: Exponential Functions & Values After 5 Years

a. Car depreciates 9% each year
Initial = $12,500
Decay rate = 0.09 → multiplier = 1 - 0.09 = 0.91
Function: C(t) = 12500 × (0.91)^t
After 5 years: C(5) = 12500 × (0.91)^5

Calculate (0.91)^5:
0.91^2 = 0.8281
0.91^4 = (0.8281)^2 ≈ 0.6857
0.91^5 = 0.6857 × 0.91 ≈ 0.6240
Then: 12500 × 0.6240 ≈ 7800

✔ Car value after 5 years ≈ $7,800

b. Baseball card increases 3% each year
Initial = $50
Growth rate = 0.03 → multiplier = 1.03
Function: B(t) = 50 × (1.03)^t
After 5 years: B(5) = 50 × (1.03)^5

Calculate (1.03)^5:
1.03^2 = 1.0609
1.03^4 = (1.0609)^2 ≈ 1.1255
1.03^5 = 1.1255 × 1.03 ≈ 1.1593
Then: 50 × 1.1593 ≈ 57.965

✔ Card value after 5 years ≈ $57.97

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Final Answer:

1.
a. Growth factor = 1.0126
b. P(x) = 6.08 × (1.0126)^x
c. Population in 2010 ≈ 6.89 billion

2.
a. V(t) = 6500 × (0.857)^t
b. Value after 3 years ≈ $4,091.10

3.
a. P(t) = 80 × (0.965)^t
b. Population drops below 15 after about 48 years

4.
a. Function: C(t) = 12500 × (0.91)^t → Value after 5 years ≈ $7,800
b. Function: B(t) = 50 × (1.03)^t → Value after 5 years ≈ $57.97