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

b. Function to model population
We use 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)

Check:
6.08 × 1.1337 = let’s compute:
6 × 1.1337 = 6.8022
0.08 × 1.1337 ≈ 0.0907
Total ≈ 6.8929 → about 6.89 billion

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Problem 2: Computer Depreciation

Given:
- Initial value = $6500
- Depreciation rate = 14.3% per year → so it retains 100% - 14.3% = 85.7% = 0.857 each year

a. Function for value
V(t) = 6500 × (0.857)^t where t = years

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

Calculate (0.857)^3:
0.857 × 0.857 = 0.734449
Then × 0.857 ≈ 0.6294 (let me check with more precision)
Actually:
0.857^2 = 0.734449
0.734449 × 0.857 = let's compute:
0.734449 × 0.8 = 0.5875592
0.734449 × 0.057 = approx 0.041863593
Sum ≈ 0.629422793 → ~0.6294

Now: 6500 × 0.6294 = ?
6500 × 0.6 = 3900
6500 × 0.0294 = 6500 × 0.03 = 195 minus 6500×0.0006=3.9 → 195 - 3.9 = 191.1? Wait no — better way:

Actually: 6500 × 0.6294
Breakdown:
6000 × 0.6294 = 3776.4
500 × 0.6294 = 314.7
Total = 3776.4 + 314.7 = 4091.1

Wait — that can’t be right because depreciation should reduce value significantly. Let me recalculate (0.857)^3 properly.

Use calculator-style steps:

0.857 × 0.857 =
857 × 857 = (800+57)^2 = 800² + 2×800×57 + 57² = 640000 + 91200 + 3249 = 734449 → so 0.734449

Then 0.734449 × 0.857:

Compute 734449 × 857 / 10^9 (since both are scaled by 10^-3 and 10^-3 again? Actually easier:

0.734449 × 0.857 =
= 0.734449 × (0.8 + 0.05 + 0.007)
= 0.734449×0.8 = 0.5875592
+ 0.734449×0.05 = 0.03672245
+ 0.734449×0.007 = 0.005141143
Sum = 0.5875592 + 0.03672245 = 0.62428165 + 0.005141143 = 0.629422793

So yes, ~0.6294

Now 6500 × 0.629422793 =
6500 × 0.629422793
= 6500 × 0.629422793
Let me do 6500 × 0.6294 = as above ~4091.1
But let’s use exact:
6500 × 0.629422793 =
First, 6500 × 0.6 = 3900
6500 × 0.029 = 188.5
6500 × 0.000422793 ≈ 6500 × 0.0004 = 2.6, plus 6500×0.000022793≈0.148 → total ~2.748
So 3900 + 188.5 = 4088.5 + 2.748 ≈ 4091.25

But wait — this seems too high? Let me verify with direct multiplication:

6500 × 0.629422793 =
Do 6500 × 629422793 / 1000000000
Too messy — use approximate:

Actually, 0.857^3 = 0.629422793
6500 × 0.629422793 = 4091.2481545 → so approximately $4091.25

But let’s round to nearest dollar: $4091

Wait — I think I made a mistake in interpretation. The depreciation is 14.3% per year, meaning each year it loses 14.3% of its current value, so multiplier is 1 - 0.143 = 0.857 — correct.

After 3 years: 6500 * (0.857)^3 ≈ 6500 * 0.6294 = 4091.1 → yes.

But let me double-check with another method:

Year 0: 6500
Year 1: 6500 * 0.857 = 5570.5
Year 2: 5570.5 * 0.857 = let's compute:
5570.5 × 0.8 = 4456.4
5570.5 × 0.057 = 317.5185
Sum = 4456.4 + 317.5185 = 4773.9185
Year 3: 4773.9185 × 0.857
4773.9185 × 0.8 = 3819.1348
4773.9185 × 0.057 = 272.1133545
Sum = 3819.1348 + 272.1133545 = 4091.2481545 → same as before.

So value after 3 years is $4091.25, or rounded to nearest cent, but since it's money, we can say $4091.25 or perhaps the problem expects whole dollars? But let's keep two decimals.

Actually, in context, maybe round to nearest dollar: $4091

But I'll write as calculated.

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Problem 3: Animal Population Decrease

Given:
- Initial count = 80 animals
- Decrease rate = 3.5% per year → so retention rate = 100% - 3.5% = 96.5% = 0.965

a. Function for population
A(t) = 80 × (0.965)^t where t = years

b. Graph and estimate when drops below 15

We need to find smallest integer t such that A(t) < 15

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

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

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

ln(0.965) ≈ ?
Using calculator: ln(0.965) ≈ -0.0356 (approx)
ln(0.1875) ≈ ln(3/16) = ln(3) - ln(16) ≈ 1.0986 - 2.7726 = -1.674

More accurately:
ln(0.1875) = ln(1875/10000) = but better use known values or calc.

Assume:
ln(0.965) ≈ -0.03562
ln(0.1875) ≈ -1.67398

So:
t > (-1.67398) / (-0.03562) ≈ 47.00 (approximately)

Calculate:
1.67398 / 0.03562 ≈ ?

0.03562 × 47 = 0.03562 × 40 = 1.4248, 0.03562 × 7 = 0.24934, sum = 1.67414 → very close to 1.67398

So t ≈ 47

Check at t=47:
A(47) = 80 × (0.965)^47

Compute (0.965)^47:
Since we have t ≈ 47 from logs, and it was very close, likely at t=47 it's just below 15.

Verify with calculation:

We know (0.965)^47 ≈ e^{47 * ln(0.965)} ≈ e^{47 * (-0.03562)} = e^{-1.67414} ≈ 0.1874 (since e^{-1.674} ≈ 0.1875)

Then 80 × 0.1874 = 14.992 < 15 → yes!

At t=46:
(0.965)^46 = (0.965)^47 / 0.965 ≈ 0.1874 / 0.965 ≈ 0.1942
80 × 0.1942 = 15.536 > 15

So at t=46, pop ≈15.54 >15
At t=47, pop≈14.99<15

Thus, first drops below 15 at year 47

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Problem 4: Exponential Functions

a. Car depreciates 9% each year → retains 91% = 0.91
Initial value = $12,500
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 ≈ 0.6240321451

Then 12500 × 0.6240321451 ≈
12500 × 0.624 = 12500 × 0.6 = 7500, 12500 × 0.024 = 300, total 7800
More precisely: 12500 × 0.6240321451 =
12500 × 0.6240321451 = let's compute 12500 × 0.6240321451
= (12500 × 0.6240321451)
Or: 12500 × 0.6240321451 = 7800.40181375 → so $7800.40

b. Baseball card increases 3% each year → growth factor = 1.03
Initial = $50
Function: V(t) = 50 × (1.03)^t
After 5 years: V(5) = 50 × (1.03)^5

(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

Then 50 × 1.1592740743 ≈ 57.963703715 → $57.96

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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. A(t) = 80 × (0.965)^t
b. Drops below 15 at year 47

Problem 4:
a. Function: V(t) = 12500 × (0.91)^t; after 5 years: $7800.40
b. Function: 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. A(t) = 80 × (0.965)^t
3b. 47 years
4a. Function: V(t) = 12500 × (0.91)^t; value after 5 years: $7800.40
4b. Function: V(t) = 50 × (1.03)^t; value after 5 years: $57.96