Worksheet featuring four word problems on exponential growth and decay, including population increase, computer depreciation, animal population decline, and value changes of a car and baseball card.

Let's solve each problem step by step, explaining the reasoning and calculations involved.

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Problem 1: World Population Growth

Given:
- World population in 2000: 6.08 billion
- Annual growth rate: 1.26%

#### a. Find the growth factor for the world population.

The growth factor is calculated as:

$$
\text{Growth factor} = 1 + \frac{\text{rate}}{100} = 1 + \frac{1.26}{100} = 1 + 0.0126 = 1.0126
$$

✔ Answer: The growth factor is 1.0126

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#### b. Write a function to model the world population.

We use the exponential growth formula:

$$
P(x) = P_0 \cdot (1 + r)^x
$$

Where:
- $ P_0 = 6.08 $ billion (initial population)
- $ r = 0.0126 $
- $ x $ = number of years after 2000

So:

$$
P(x) = 6.08 \cdot (1.0126)^x
$$

✔ Answer: $ P(x) = 6.08 \cdot (1.0126)^x $

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#### c. Find the world population in 2010.

In 2010, $ x = 10 $ years after 2000.

$$
P(10) = 6.08 \cdot (1.0126)^{10}
$$

First calculate $ (1.0126)^{10} $:

Using a calculator:
$$
(1.0126)^{10} \approx 1.1354
$$

Now:
$$
P(10) \approx 6.08 \cdot 1.1354 \approx 6.895 \text{ billion}
$$

✔ Answer: The world population in 2010 is approximately 6.895 billion

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Problem 2: Computer Depreciation

Given:
- Initial value: $6500
- Depreciates at 14.3% per year

#### a. Write a function that models the value of the computer.

Depreciation means decay, so we use:

$$
V(x) = V_0 \cdot (1 - r)^x
$$

Where:
- $ V_0 = 6500 $
- $ r = 0.143 $
- $ x $ = number of years

So:

$$
V(x) = 6500 \cdot (1 - 0.143)^x = 6500 \cdot (0.857)^x
$$

✔ Answer: $ V(x) = 6500 \cdot (0.857)^x $

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#### b. Find the value after three years

$$
V(3) = 6500 \cdot (0.857)^3
$$

Calculate $ (0.857)^3 $:

$$
0.857^3 \approx 0.630
$$

Then:
$$
V(3) \approx 6500 \cdot 0.630 = 4095
$$

✔ Answer: The value after 3 years is approximately $4,095

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Problem 3: Animal Population Decline

Given:
- Initial population: 80 animals
- Decreases at 3.5% per year

#### a. Write a function that models the change in animal population.

Use decay model:

$$
A(x) = A_0 \cdot (1 - r)^x
$$

Where:
- $ A_0 = 80 $
- $ r = 0.035 $
- $ x $ = number of years

So:

$$
A(x) = 80 \cdot (1 - 0.035)^x = 80 \cdot (0.965)^x
$$

✔ Answer: $ A(x) = 80 \cdot (0.965)^x $

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#### b. Graph the function. Estimate the number of years until the population drops below 15 animals.

We need to find the smallest integer $ x $ such that:

$$
80 \cdot (0.965)^x < 15
$$

Divide both sides by 80:

$$
(0.965)^x < \frac{15}{80} = 0.1875
$$

Take logarithm of both sides:

$$
\log((0.965)^x) < \log(0.1875)
\Rightarrow x \cdot \log(0.965) < \log(0.1875)
$$

Note: $ \log(0.965) $ is negative, so when dividing, reverse inequality:

$$
x > \frac{\log(0.1875)}{\log(0.965)}
$$

Calculate:

- $ \log(0.1875) \approx -0.724 $
- $ \log(0.965) \approx -0.0156 $

$$
x > \frac{-0.724}{-0.0156} \approx 46.4
$$

So $ x \geq 47 $

✔ Answer: It takes about 47 years for the population to drop below 15 animals.

> Note: You can graph $ A(x) = 80 \cdot (0.965)^x $ and find where it crosses 15. The graph will be decreasing exponentially.

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Problem 4: Write Exponential Functions and Evaluate After 5 Years

#### a. A $12,500 car depreciates 9% each year

This is decay:

$$
V(x) = 12500 \cdot (1 - 0.09)^x = 12500 \cdot (0.91)^x
$$

After 5 years:

$$
V(5) = 12500 \cdot (0.91)^5
$$

Calculate $ (0.91)^5 $:

$$
(0.91)^5 \approx 0.6497
$$

$$
V(5) \approx 12500 \cdot 0.6497 \approx 8121.25
$$

✔ Answer: Function: $ V(x) = 12500 \cdot (0.91)^x $; Value after 5 years ≈ $8,121.25

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#### b. A baseball card bought for $50 increases 3% in value each year

This is growth:

$$
V(x) = 50 \cdot (1 + 0.03)^x = 50 \cdot (1.03)^x
$$

After 5 years:

$$
V(5) = 50 \cdot (1.03)^5
$$

Calculate $ (1.03)^5 \approx 1.1593 $

$$
V(5) \approx 50 \cdot 1.1593 \approx 57.965
$$

✔ Answer: Function: $ V(x) = 50 \cdot (1.03)^x $; Value after 5 years ≈ $57.97

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✔ Final Answers Summary

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1. World Population
- a. Growth factor: 1.0126
- b. Function: $ P(x) = 6.08 \cdot (1.0126)^x $
- c. Population in 2010: ≈ 6.895 billion

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2. Computer Depreciation
- a. Function: $ V(x) = 6500 \cdot (0.857)^x $
- b. Value after 3 years: ≈ $4,095

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3. Animal Population
- a. Function: $ A(x) = 80 \cdot (0.965)^x $
- b. Drops below 15 in ≈ 47 years

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4. Exponential Models
- a. Car: $ V(x) = 12500 \cdot (0.91)^x $ → $ V(5) ≈ \$8,121.25 $
- b. Baseball card: $ V(x) = 50 \cdot (1.03)^x $ → $ V(5) ≈ \$57.97 $

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