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

Let’s solve each problem one by one, step by step. We’ll use simple math and logic — no fancy terms needed!

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Problem 1:
> In 2000, the world population was 6 billion. The annual rate of growth is about 1%. If this rate continues to be constant:
> a) How many people will there be in 2050? Write a function to model this.
> b) What is the number of people per year after 2000? Find the exact equation for it.

We’re dealing with exponential growth here — that means the population grows by a percentage every year, not a fixed number.

The general formula for exponential growth is:

> P(t) = P₀ × (1 + r)^t

Where:
- P(t) = population after t years
- P₀ = starting population (in 2000)
- r = growth rate per year (as a decimal)
- t = number of years since 2000

Given:
- P₀ = 6 billion = 6,000,000,000
- r = 1% = 0.01
- For part (a), t = 2050 - 2000 = 50 years

So,

P(50) = 6,000,000,000 × (1 + 0.01)^50
= 6,000,000,000 × (1.01)^50

Now calculate (1.01)^50:

Using calculator:
1.01^50 ≈ 1.64463

Then:

P(50) ≈ 6,000,000,000 × 1.64463 ≈ 9,867,780,000

So about 9.87 billion people in 2050.

Function to model:
P(t) = 6,000,000,000 × (1.01)^t

Part (b): “What is the number of people per year after 2000?” — This seems to ask for the same function, just written clearly.

Exact equation:
P(t) = 6 × 10⁹ × (1.01)^t (using scientific notation)

Or if they want it without exponents in base 10:
P(t) = 6,000,000,000 × (1.01)^t

✔ So we’ve got both parts.

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Problem 2:
> A computer valued at $4000 depreciates at a rate of 30% per year.
> a) Write a function that models the value of the computer over time.
> b) Find the value of the computer after three years.

Depreciation means losing value — so instead of growing, it’s shrinking by 30% each year.

That means each year, it keeps 70% of its value (because 100% - 30% = 70%).

Formula for decay:

> V(t) = V₀ × (1 - r)^t

Where:
- V(t) = value after t years
- V₀ = initial value = $4000
- r = depreciation rate = 30% = 0.30
- So (1 - r) = 0.70

Function:
V(t) = 4000 × (0.70)^t

After 3 years: t = 3

V(3) = 4000 × (0.70)^3
Calculate (0.70)^3 = 0.7 × 0.7 × 0.7 = 0.343

Then:
V(3) = 4000 × 0.343 = 1372

So value after 3 years = $1,372

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Problem 3:
> The population of a certain animal species decreases at a rate of 3% per year. You have counted 80 of the animals in the habitat you are studying.
> a) Write a function to model the population decrease.
> b) Graph the function. Estimate the length of years until the population drops below 15 animals.

Again, exponential decay.

Rate = 3% → so multiplier = 1 - 0.03 = 0.97

Initial population = 80

Function:
P(t) = 80 × (0.97)^t

Part (b): We need to find when P(t) < 15

So solve:
80 × (0.97)^t < 15

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

Take log of both sides (or guess and check):

Try t = 50:
0.97^50 ≈ ? Let’s compute step by step or use approx.

Actually, let’s do trial values:

t = 50: 0.97^50 ≈ e^(50 * ln(0.97)) ≈ e^(50 * -0.03046) ≈ e^(-1.523) ≈ 0.218 → too high

t = 60: 0.97^60 ≈ e^(60 * -0.03046) ≈ e^(-1.8276) ≈ 0.160 → still above 0.1875? Wait no — 0.160 < 0.1875? No, 0.160 is less than 0.1875 → wait, we want (0.97)^t < 0.1875

Wait — 0.97^50 ≈ 0.218 > 0.1875 → not yet

0.97^55: let’s try 55

ln(0.97) ≈ -0.03046
55 × -0.03046 = -1.6753
e^-1.6753 ≈ 0.187 → very close!

So at t = 55, (0.97)^55 ≈ 0.187 → then P(55) = 80 × 0.187 ≈ 14.96 → which is under 15!

Check t = 54:

54 × -0.03046 = -1.64484
e^-1.64484 ≈ 0.193
P(54) = 80 × 0.193 ≈ 15.44 → still above 15

So between 54 and 55 years.

Since we can’t have half a year usually, and question says “estimate”, we say about 55 years.

Graphing: It would be a curve starting at 80 and slowly dropping toward zero, getting flatter as time goes on.

But since we don’t graph here, we just estimate the time.

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Problem 4:
> Write an exponential function to match each situation. Find the value at least two times.
> a) $1000 invested at 4% interest compounded annually.
> b) A car bought for $20,000 loses 15% of its value each year.

Part (a): Investment growing at 4% per year.

Growth formula: A(t) = P × (1 + r)^t
P = 1000, r = 0.04

A(t) = 1000 × (1.04)^t

Find value at two times — let’s pick t=1 and t=5

t=1: A(1) = 1000 × 1.04 = $1040
t=5: A(5) = 1000 × (1.04)^5 ≈ 1000 × 1.21665 ≈ $1216.65

Part (b): Car depreciating 15% per year → keeps 85%

V(t) = 20000 × (0.85)^t

Pick t=1 and t=3

t=1: V(1) = 20000 × 0.85 = $17,000
t=3: V(3) = 20000 × (0.85)^3 = 20000 × 0.614125 = $12,282.50

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Now let’s write all final answers neatly.

Final Answer:

1a) Population in 2050: approximately 9,867,780,000 (or 9.87 billion). Function: P(t) = 6,000,000,000 × (1.01)^t
1b) Exact equation: P(t) = 6 × 10⁹ × (1.01)^t

2a) Value function: V(t) = 4000 × (0.70)^t
2b) Value after 3 years: $1,372

3a) Population function: P(t) = 80 × (0.97)^t
3b) Population drops below 15 after about 55 years.

4a) Investment function: A(t) = 1000 × (1.04)^t; at t=1: $1040, at t=5: ~$1216.65
4b) Car value function: V(t) = 20000 × (0.85)^t; at t=1: $17,000, at t=3: $12,282.50