20 Exponential Growth and Decay Word Problems | PDF - Free Printable

Let’s solve each problem step by step.

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Problem 1: World Population Growth

Given:
- Initial population (in 2000) = 6.08 billion
- Annual growth rate = 1.26% → which is 0.0126 in decimal

a. Find the growth factor.

Growth factor = 1 + growth rate
= 1 + 0.0126
= 1.0126

✔ Checked: Correct.

b. Write a function to model the world population.

Let x = number of years past 2000
Population P(x) = initial × (growth factor)^x
P(x) = 6.08 × (1.0126)^x

✔ Checked: Standard exponential growth formula.

c. Find population in 2010.

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

Calculate (1.0126)^10:

Using calculator:
1.0126^10 ≈ 1.1327 (rounded to 4 decimals)

Then:
6.08 × 1.1327 ≈ ?

6 × 1.1327 = 6.7962
0.08 × 1.1327 ≈ 0.0906
Total ≈ 6.7962 + 0.0906 = 6.8868 billion

More accurately:
6.08 × 1.1327 = let’s compute:

6.08 × 1.1327
= 6.08 × (1 + 0.1327)
= 6.08 + 6.08×0.1327
6.08 × 0.13 = 0.7904
6.08 × 0.0027 = ~0.0164
So 0.7904 + 0.0164 = 0.8068
Total = 6.08 + 0.8068 = 6.8868 billion

Rounded to two decimal places: 6.89 billion

✔ Checked: Reasonable estimate.

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Problem 2: Computer Depreciation

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

a. Write a function that models the value.

Depreciation means we multiply by (1 - decay rate) each year.

Decay factor = 1 - 0.143 = 0.857

Let t = number of years
V(t) = 6500 × (0.857)^t

✔ Checked: Correct for exponential decay.

b. Value after three years.

V(3) = 6500 × (0.857)^3

First, calculate 0.857^3:

0.857 × 0.857 = ?
0.857 × 0.857:
= (0.8 + 0.057)^2 = 0.64 + 2×0.8×0.057 + (0.057)^2
But easier with direct multiplication:

0.857 × 0.857:
857 × 857 = ?

Actually, use calculator-style:

0.857 × 0.857 = 0.734449
Then × 0.857 again:

0.734449 × 0.857 ≈ ?

0.734449 × 0.8 = 0.5875592
0.734449 × 0.057 = approx 0.041863593
Sum ≈ 0.5875592 + 0.041863593 = 0.629422793

So V(3) = 6500 × 0.629422793 ≈ ?

6500 × 0.6294 ≈ ?

6000 × 0.6294 = 3776.4
500 × 0.6294 = 314.7
Total = 3776.4 + 314.7 = 4091.1

More accurately:
6500 × 0.629422793 = let’s compute:

6500 × 0.629422793
= 6500 × 0.629422793
≈ 4091.248

So approximately $4091.25

✔ Checked: Makes sense — lost about 37% over 3 years.

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Problem 3: Animal Population Decrease

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

a. Write a function modeling the change.

Decay factor = 1 - 0.035 = 0.965

Let t = number of years
P(t) = 80 × (0.965)^t

✔ Checked: Correct.

b. Graph the function. Estimate when population drops below 15.

We need to find smallest integer t such that:

80 × (0.965)^t < 15

Divide both sides by 80:

(0.965)^t < 15/80 = 0.1875

Take natural log or common log on both sides:

ln((0.965)^t) < ln(0.1875)
t × ln(0.965) < ln(0.1875)

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

Compute:

ln(0.1875) ≈ -1.674
ln(0.965) ≈ -0.0356

So:

t > (-1.674) / (-0.0356) ≈ 47.02

So t ≈ 47 years

Check at t=47:

P(47) = 80 × (0.965)^47

Compute (0.965)^47:

Use approximation:
(0.965)^47 = e^(47 × ln(0.965)) ≈ e^(47 × -0.0356) ≈ e^(-1.6732) ≈ 0.1875

Exactly what we had! So P(47) = 80 × 0.1875 = 15 → exactly 15

We want *below* 15 → so try t=48

P(48) = 80 × (0.965)^48 = 80 × 0.965 × 0.1875 ≈ 80 × 0.1809375 ≈ 14.475

Yes, below 15.

So answer: 48 years

✔ Checked: At t=47, it's exactly 15; at t=48, it’s below.

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Problem 4: Write exponential functions and find value after 5 years

a. Car depreciates 9% each year, initial $12,500

Depreciation rate = 9% → decay factor = 1 - 0.09 = 0.91

Function: V(t) = 12500 × (0.91)^t

After 5 years: V(5) = 12500 × (0.91)^5

Compute (0.91)^5:

0.91^2 = 0.8281
0.91^4 = (0.8281)^2 = 0.68574961
0.91^5 = 0.68574961 × 0.91 ≈ ?

0.68574961 × 0.9 = 0.617174649
0.68574961 × 0.01 = 0.0068574961
Sum ≈ 0.6240321451

So V(5) = 12500 × 0.6240321451 ≈ ?

12500 × 0.624 = 12500 × 0.6 = 7500; 12500 × 0.024 = 300 → total 7800
More precisely:

12500 × 0.6240321451 =
= 12500 × 0.6240321451
≈ 7800.4018

So ≈ $7800.40

✔ Checked.

b. Baseball card bought for $50, increases 3% each year

Growth rate = 3% → growth factor = 1.03

Function: V(t) = 50 × (1.03)^t

After 5 years: V(5) = 50 × (1.03)^5

Compute (1.03)^5:

1.03^2 = 1.0609
1.03^4 = (1.0609)^2 = 1.12550881
1.03^5 = 1.12550881 × 1.03 ≈ ?

1.12550881 × 1 = 1.12550881
1.12550881 × 0.03 = 0.0337652643
Sum = 1.1592740743

So V(5) = 50 × 1.1592740743 ≈ 57.9637

≈ $57.96

✔ Checked.

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Final Answers:

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

Problem 2:
a. V(t) = 6500 × (0.857)^t
b. Value after 3 years ≈ $4091.25

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

Problem 4:
a. V(t) = 12500 × (0.91)^t; after 5 years ≈ $7800.40
b. V(t) = 50 × (1.03)^t; after 5 years ≈ $57.96

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Final Answer:
1a. 1.0126
1b. P(x) = 6.08 × (1.0126)^x
1c. 6.89 billion
2a. V(t) = 6500 × (0.857)^t
2b. $4091.25
3a. P(t) = 80 × (0.965)^t
3b. 48 years
4a. V(t) = 12500 × (0.91)^t; $7800.40
4b. V(t) = 50 × (1.03)^t; $57.96