Exponential Growth And Decay Word Problems Worksheet - 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 (as a decimal)

a. Find the growth factor

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

✔ So, the growth factor is 1.0126

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b. Write a function to model the world population

We use exponential growth formula:

> P(x) = Initial amount × (growth factor)^x

Where:
- x = number of years after 2000
- Initial amount = 6.08 billion
- Growth factor = 1.0126

So,

> P(x) = 6.08 × (1.0126)^x

This function gives population in billions, x years after 2000.

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c. Find population in 2010

2010 is 10 years after 2000 → x = 10

Plug into the function:

P(10) = 6.08 × (1.0126)^10

First, calculate (1.0126)^10:

Using calculator:
1.0126^10 ≈ 1.1335 (rounded to 4 decimals)

Now multiply:

6.08 × 1.1335 ≈ ?

6.08 × 1.1335 = let’s compute:

6 × 1.1335 = 6.801
0.08 × 1.1335 = 0.09068
Total ≈ 6.801 + 0.09068 = 6.89168

So, approximately 6.89 billion

✔ Final answer for part c: About 6.89 billion

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Problem 2: Computer Depreciation

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

Depreciation means value decreases → we use decay factor.

Decay factor = 1 - depreciation rate = 1 - 0.143 = 0.857

a. Write a function that models the value

Use exponential decay:

> V(t) = Initial value × (decay factor)^t

Where t = number of years

So,

> V(t) = 6500 × (0.857)^t

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b. Value after 3 years

t = 3

V(3) = 6500 × (0.857)^3

First, calculate (0.857)^3:

0.857 × 0.857 = 0.734449
Then × 0.857 ≈ 0.734449 × 0.857 ≈ let’s compute:

0.734449 × 0.8 = 0.5875592
0.734449 × 0.057 = approx 0.0418636
Add them: 0.5875592 + 0.0418636 ≈ 0.6294228

So, (0.857)^3 ≈ 0.6294

Now multiply by 6500:

6500 × 0.6294 = ?

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

✔ So, value after 3 years ≈ $4091.10

(Rounded to nearest cent)

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Problem 3: Animal Population Decrease

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

Decay factor = 1 - 0.035 = 0.965

a. Write a function modeling the change

Let t = number of years

> A(t) = 80 × (0.965)^t

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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

Now take logarithm or try values of t until it drops below 0.1875.

Try t = 40:

0.965^40 = ? Let’s compute step by step or use approximation.

Actually, better to use log:

Take natural 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 inequality flips when dividing.

Compute:

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

So:

t > (-1.674) / (-0.0356) ≈ 47.02

So, t ≈ 47.02 → meaning at t = 48, it will be below 15? Wait — let’s check with actual values to be sure.

Better to test integer values around 47.

Check t = 47:

A(47) = 80 × (0.965)^47

Calculate (0.965)^47:

Use calculator: 0.965^47 ≈ ?

Or approximate:

We know from above: t ≈ 47.02 → so at t=47, still slightly above 15? Let’s compute numerically.

Actually, let’s do this carefully.

Use calculator steps:

0.965^47:

First, 0.965^10 ≈ 0.698 (approx)
But better to use direct calc:

Assume we have calculator:

0.965^47 ≈ e^(47 * ln(0.965))
ln(0.965) ≈ -0.03561
47 × -0.03561 ≈ -1.67367
e^(-1.67367) ≈ 0.1875 (wait, that’s exactly our target!)

Wait — earlier we had:

(0.965)^t = 0.1875 → t = ln(0.1875)/ln(0.965) ≈ (-1.674)/(-0.03561) ≈ 47.01

So at t = 47.01, population = 15

Therefore, at t = 47, population is just above 15
At t = 48, it’s below 15

Let’s verify:

Compute A(47) = 80 × (0.965)^47

Since (0.965)^47 ≈ 0.1875 → 80 × 0.1875 = 15 → exactly at t=47.01

So at t=47, it’s very close to 15, maybe 15.01 or something.

But since we want “first drops below 15”, we need t where A(t) < 15.

Try t=47:

Compute more accurately:

Use calculator:

0.965^47 = ?

Let me simulate:

0.965^10 = 0.6983
0.965^20 = (0.6983)^2 ≈ 0.4876
0.965^40 = (0.4876)^2 ≈ 0.2377
Now 0.965^47 = 0.965^40 × 0.965^7

0.965^7 ≈ ?
0.965^2 = 0.931225
0.965^4 = (0.931225)^2 ≈ 0.8672
0.965^6 = 0.8672 × 0.931225 ≈ 0.8075
0.965^7 = 0.8075 × 0.965 ≈ 0.7792

So 0.965^47 ≈ 0.2377 × 0.7792 ≈ 0.1852

Then A(47) = 80 × 0.1852 ≈ 14.816 → which is BELOW 15!

Wait, that contradicts earlier log calculation.

I think I made an error in manual exponentiation.

Better to trust calculator or precise computation.

Actually, let’s use exact calculation:

Using calculator:

0.965^47 = 0.1852 (approx) → 80 × 0.1852 = 14.816 → less than 15

But according to log method, t=47.01 should give 15.

Inconsistency due to rounding.

Let me recalculate log:

ln(0.1875) = ln(3/16) = ln(3) - ln(16) ≈ 1.0986 - 2.7726 = -1.674

ln(0.965) = ? Using better precision:

0.965 → ln(0.965) = -0.035610 (more precisely)

So t = (-1.674) / (-0.035610) ≈ 47.008

So at t=47.008, population = 15

Thus, at t=47, population is still above 15? Or below?

If t=47.008 is when it hits 15, then at t=47, it's slightly above 15.

But my manual calc gave 14.816 — which suggests error in manual exponentiation.

Let me use a different approach.

Use iterative multiplication or known tool.

Actually, let’s accept the log result: t ≈ 47.01

So population reaches 15 at about 47.01 years.

Therefore, it first drops below 15 during the 48th year? No — at t=47.01, it’s 15, so for any t>47.01, it’s below.

But since t is continuous, and we’re asked for "number of years until", probably they expect integer year when it first goes below.

So check t=47 and t=48.

Compute A(47) = 80 * (0.965)^47

Using precise calculator:

0.965^47 = 0.1852 (let’s say) → 80*0.1852=14.816 <15 → so at t=47, already below?

That can’t be if log says 47.01.

Perhaps my manual calc of 0.965^47 is wrong.

Let me calculate 0.965^47 properly.

Use: 0.965^47 = e^{47 * ln(0.965)}

ln(0.965) = -0.035610 (from calculator)

47 * -0.035610 = -1.67367

e^{-1.67367} = 1 / e^{1.67367}

e^1.67367 ≈ ? e^1.6 = 4.953, e^1.7=5.473, so interpolate.

1.67367 - 1.6 = 0.07367

Difference 0.1 in exponent → difference in value ~0.52, so per 0.01 is 0.052

So e^1.67367 ≈ 4.953 + 0.7367*0.52 ≈ wait, better:

Actually, e^1.67367 ≈ 5.332 (using calculator)

So e^{-1.67367} ≈ 1/5.332 ≈ 0.1875

Yes! So (0.965)^47 = 0.1875

Then A(47) = 80 * 0.1875 = 15 exactly

Oh! So at t=47, population is exactly 15.

The question says "drops below 15", so we need when it is less than 15.

So at t=47, it is 15, not below.

At t=48:

A(48) = 80 * (0.965)^48 = 80 * 0.965 * (0.965)^47 = 80 * 0.965 * 0.1875

Compute: 0.965 * 0.1875 = 0.1809375

Then 80 * 0.1809375 = 14.475

Which is below 15.

So at t=48, population is 14.475 < 15.

Therefore, it first drops below 15 at t=48 years.

✔ Answer for 3b: 48 years

(Note: If the problem allows fractional years, it would be just over 47, but since it says "estimate the number of years", and typically we report whole years when it first happens, 48 is correct.)

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Final Answers:

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

Problem 2:
a. V(t) = 6500 × (0.857)^t
b. $4091.10

Problem 3:
a. A(t) = 80 × (0.965)^t
b. 48 years

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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. $4091.10
3a. A(t) = 80 × (0.965)^t
3b. 48