Worksheet featuring four word problems on exponential growth and decay, including population changes, depreciation, and value increases.

Let’s solve each problem step by step. We’ll use the exponential growth and decay formula:

> For growth:
> `A = P(1 + r)^t`
> where:
> - A = final amount
> - P = initial amount
> - r = rate (as a decimal)
> - t = time in years

> For decay:
> `A = P(1 - r)^t`

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Problem 1: World Population

Given:
- Initial population in 2000: 6.08 billion
- Annual increase rate: 1.26% → as decimal: 0.0126

#### a. Growth factor
Growth factor = 1 + r = 1 + 0.0126 = 1.0126

✔ Check: Yes, for growth, we add the rate to 1.

#### b. Function to model population
Let x = number of years after 2000.
Population function:
P(x) = 6.08 × (1.0126)^x

✔ This matches the standard form: initial × (growth factor)^time

#### 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 ≈ 1.1335 (let me verify:
1.0126^2 = 1.02535876
^4 = (1.02535876)^2 ≈ 1.05136
^8 ≈ (1.05136)^2 ≈ 1.10536
Then ^10 = ^8 × ^2 ≈ 1.10536 × 1.02536 ≈ 1.1335 — yes)

Now multiply:
6.08 × 1.1335 ≈ ?

6 × 1.1335 = 6.801
0.08 × 1.1335 ≈ 0.09068
Total ≈ 6.89168 → about 6.89 billion

✔ Double-check with direct calc:
6.08 * (1.0126)^10 = 6.08 * 1.1335 ≈ 6.891 → rounds to 6.89 billion

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Problem 2: Computer Depreciation

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

#### a. Function for value
Decay formula: V(t) = P(1 - r)^t
V(t) = 6500 × (1 - 0.143)^t = 6500 × (0.857)^t

✔ Correct: 1 - 0.143 = 0.857

#### b. Value after 3 years
t = 3

V(3) = 6500 × (0.857)^3

Calculate (0.857)^3:

0.857 × 0.857 = 0.734449
× 0.857 ≈ 0.734449 × 0.857

Compute:
0.734449 × 0.8 = 0.5875592
0.734449 × 0.057 = approx 0.0418636
Total ≈ 0.6294228

So V(3) ≈ 6500 × 0.6294228 ≈ ?

6500 × 0.6 = 3900
6500 × 0.0294228 ≈ 6500 × 0.03 = 195, minus 6500×0.0005772≈3.75 → so ~191.25
Total ≈ 3900 + 191.25 = 4091.25

Better to compute directly:

6500 × 0.6294228 = let's do 6500 × 0.6294 = ?

6500 × 0.6 = 3900
6500 × 0.0294 = 6500 × 0.03 = 195 minus 6500×0.0006=3.9 → 191.1
So 3900 + 191.1 = 4091.1

Actual precise calculation:

0.857^3 = 0.857 * 0.857 = 0.734449; then *0.857 = 0.629422793

6500 * 0.629422793 = 4091.2481545 → $4091.25

✔ Rounded to nearest cent: $4091.25

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Problem 3: Animal Population Decrease

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

#### a. Function for population
P(t) = 80 × (1 - 0.035)^t = 80 × (0.965)^t

✔ Correct decay model.

#### b. Graph and 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 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 when dividing, inequality flips.

t > log(0.1875) / log(0.965)

Calculate:

log(0.1875) ≈ -0.727 (using base 10)
log(0.965) ≈ -0.0154 (approx)

So t > (-0.727) / (-0.0154) ≈ 47.2

So around 48 years

Check with actual values:

Try t = 47:

0.965^47 = ? Let’s compute step by step or use approximation.

Alternatively, use calculator-style steps:

We can compute iteratively or use known values.

But since this is estimation via graphing, and we got t > 47.2, so first full year it drops below 15 is at t = 48

Verify:

At t=47: 80*(0.965)^47

Compute (0.965)^47:

Use ln: ln(0.965) ≈ -0.0356
47 * (-0.0356) = -1.6732
e^{-1.6732} ≈ 0.1877 → 80*0.1877 ≈ 15.016 → still above 15

At t=48: 48 * (-0.0356) = -1.7088
e^{-1.7088} ≈ 0.1812 → 80*0.1812 ≈ 14.496 → below 15

✔ So answer: 48 years

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Problem 4: Write functions and find value after 5 years

#### a. Car depreciates 9% each year

Initial: $12,500
Rate: 9% → r = 0.09
Decay: V(t) = 12500 × (1 - 0.09)^t = 12500 × (0.91)^t

After 5 years: t=5

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 ≈ 0.6240321451

Now: 12500 × 0.6240321451 ≈ ?

12500 × 0.6 = 7500
12500 × 0.024 = 300
12500 × 0.0000321451 ≈ 0.4018
Total ≈ 7500 + 300 = 7800 + 0.40 ≈ 7800.40? Wait no:

Wait: 0.6240321451 × 12500

Better: 12500 × 0.6240321451 = 12500 × 0.624 = 7800, plus 12500×0.0000321451≈0.4018 → but actually:

0.6240321451 × 12500 = let's compute:

= (0.6240321451 × 10000) + (0.6240321451 × 2500)
= 6240.321451 + 1560.08036275 = 7800.40181375

So approximately $7800.40

✔ Final value: $7800.40

#### b. Baseball card increases 3% each year

Initial: $50
Rate: 3% → r = 0.03
Growth: V(t) = 50 × (1 + 0.03)^t = 50 × (1.03)^t

After 5 years: t=5

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 ≈ 1.1592740743

Now: 50 × 1.1592740743 ≈ 57.963703715 → $57.96

✔ Final value: $57.96

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Final Answer:

1.
a. 1.0126
b. P(x) = 6.08 × (1.0126)^x
c. Approximately 6.89 billion

2.
a. V(t) = 6500 × (0.857)^t
b. $4091.25

3.
a. P(t) = 80 × (0.965)^t
b. 48 years

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