Algebra 2 Unit 6 Test Answer Key

Algebra 2 Unit 6 tests typically cover a range of topics, including exponential and logarithmic functions, their properties, equations, and applications. This domain of mathematics is critical for understanding phenomena that grow or decay over time, making it a cornerstone in fields like finance, physics, and computer science. Mastering these concepts is essential not only for academic success in Algebra 2 but also for future studies in higher mathematics and related disciplines.

Exponential Functions: A Deep Dive

Exponential functions form the bedrock of Unit 6. These functions are defined by the equation f(x) = aˣ, where a is a constant greater than 0 and not equal to 1. The variable x appears as an exponent, which dictates the function's behavior. The value of a, known as the base, determines whether the function represents exponential growth or decay.

• Exponential Growth: When a > 1, the function increases as x increases. This models situations where a quantity grows rapidly over time, such as population growth or compound interest.
• Exponential Decay: When 0 < a < 1, the function decreases as x increases. This models situations where a quantity diminishes over time, such as radioactive decay or the depreciation of an asset.

The domain of an exponential function is all real numbers, meaning you can input any real number for x. The range, however, is all positive real numbers, indicating that the output f(x) is always positive.

Key Properties:

• The graph of f(x) = aˣ always passes through the point (0, 1) because any number raised to the power of 0 is 1.
• The x-axis is a horizontal asymptote, meaning the graph approaches the x-axis but never touches it.
• Exponential functions are one-to-one, meaning each input x corresponds to a unique output f(x), which allows for the existence of an inverse function (logarithmic function).

Examples:

• f(x) = 2ˣ (Exponential Growth)
• f(x) = (1/2)ˣ (Exponential Decay)

Logarithmic Functions: The Inverse Operation

Logarithmic functions are the inverses of exponential functions. This means that if y = aˣ, then x = logₐ(y). In simpler terms, the logarithm answers the question: "To what power must I raise a to get y?"

The logarithmic function is defined as f(x) = logₐ(x), where a is the base, x is the argument, and a > 0 and a ≠ 1. The domain of a logarithmic function is all positive real numbers, and the range is all real numbers.

Common Logarithms:

• Common Logarithm: The logarithm with base 10, denoted as log₁₀(x) or simply log(x).
• Natural Logarithm: The logarithm with base e (Euler's number, approximately 2.71828), denoted as logₑ(x) or ln(x).

Key Properties:

• logₐ(1) = 0 because a⁰ = 1
• logₐ(a) = 1 because a¹ = 1
• logₐ(aˣ) = x (Inverse Property)
• a^(logₐ(x)) = x (Inverse Property)

Laws of Logarithms: Simplifying Expressions

The laws of logarithms are crucial for simplifying and solving logarithmic equations. These laws stem directly from the properties of exponents.

1. Product Rule: logₐ(xy) = logₐ(x) + logₐ(y)

   • The logarithm of a product is the sum of the logarithms.

2. Quotient Rule: logₐ(x/y) = logₐ(x) - logₐ(y)

   • The logarithm of a quotient is the difference of the logarithms.

3. Power Rule: logₐ(xⁿ) = nlogₐ(x)*

   • The logarithm of a power is the exponent times the logarithm.

4. Change of Base Formula: logₐ(x) = logₓ(x) / logₓ(a)

   • This formula allows you to convert logarithms from one base to another, which is particularly useful when using calculators that only have common or natural logarithm functions.

Examples:

• log₂(8 * 4) = log₂(8) + log₂(4) = 3 + 2 = 5
• log₅(25 / 5) = log₅(25) - log₅(5) = 2 - 1 = 1
• log₃(9²) = 2 * log₃(9) = 2 * 2 = 4
• log₂(7) = ln(7) / ln(2) (using the change of base formula to base e)

Solving Exponential Equations

Exponential equations involve variables in the exponent. To solve these equations, the goal is to isolate the exponential term and then use logarithms to bring the exponent down.

Steps:

1. Isolate the exponential term: Get the exponential term alone on one side of the equation.
2. Take the logarithm of both sides: Apply either the common logarithm or the natural logarithm to both sides of the equation. Choose the logarithm that simplifies the equation most effectively.
3. Apply the power rule of logarithms: Use the property logₐ(xⁿ) = nlogₐ(x)* to bring the exponent down.
4. Solve for the variable: Perform algebraic manipulations to isolate the variable.

Examples:

1. Solve for x: 2ˣ = 8

   • Take the logarithm of both sides: log(2ˣ) = log(8)
   • Apply the power rule: xlog(2) = log(8)*
   • Solve for x: x = log(8) / log(2) = 3

2. Solve for x: 3^(x+1) = 27

   • Take the logarithm of both sides: log(3^(x+1)) = log(27)
   • Apply the power rule: *(x+1)log(3) = log(27)
   • Solve for x: x+1 = log(27) / log(3) = 3
   • x = 3 - 1 = 2

3. Solve for x: 5 * 2^(3x-2) = 40

   • Divide both sides by 5: 2^(3x-2) = 8
   • Take the logarithm of both sides: log(2^(3x-2)) = log(8)
   • Apply the power rule: (3x-2)log(2) = log(8)
   • Divide both sides by log(2): 3x-2 = log(8) / log(2) = 3
   • Add 2 to both sides: 3x = 5
   • Divide both sides by 3: x = 5/3

Solving Logarithmic Equations

Logarithmic equations involve variables within the argument of a logarithm. To solve these equations, the goal is to isolate the logarithmic term and then exponentiate both sides.

Steps:

1. Isolate the logarithmic term: Get the logarithmic term alone on one side of the equation. Use the laws of logarithms to combine multiple logarithmic terms into a single logarithm if necessary.
2. Exponentiate both sides: Raise the base of the logarithm to the power of both sides of the equation. This will "undo" the logarithm.
3. Solve for the variable: Perform algebraic manipulations to isolate the variable.
4. Check for extraneous solutions: Because the domain of a logarithmic function is all positive real numbers, it's crucial to check that the solutions you find do not result in taking the logarithm of a negative number or zero in the original equation.

Examples:

1. Solve for x: log₂(x) = 3

   • Exponentiate both sides: 2^(log₂(x)) = 2³
   • Simplify: x = 8

2. Solve for x: log(x + 2) + log(x - 1) = 1

   • Combine the logarithms: log((x + 2)(x - 1)) = 1
   • Exponentiate both sides: 10^(log((x + 2)(x - 1))) = 10¹
   • Simplify: (x + 2)(x - 1) = 10
   • Expand and solve the quadratic equation: x² + x - 2 = 10
   • x² + x - 12 = 0
   • (x + 4)(x - 3) = 0
   • x = -4 or x = 3
   • Check for extraneous solutions:
     • If x = -4, log(-4 + 2) = log(-2), which is undefined. So, x = -4 is an extraneous solution.
     • If x = 3, log(3 + 2) + log(3 - 1) = log(5) + log(2) = log(10) = 1. So, x = 3 is a valid solution.

3. Solve for x: ln(5x - 1) = ln(3x + 5)

   • Since the logarithms are equal, their arguments must be equal: 5x - 1 = 3x + 5
   • Solve for x: 2x = 6
   • x = 3
   • Check for extraneous solutions:
     • ln(5(3) - 1) = ln(14) and ln(3(3) + 5) = ln(14). The solution is valid.

Applications of Exponential and Logarithmic Functions

Exponential and logarithmic functions have a wide array of applications in various fields:

• Finance: Compound interest, present value calculations, and loan amortization.
• Biology: Population growth, radioactive decay, and enzyme kinetics.
• Physics: Newton's law of cooling, sound intensity (decibels), and electrical circuits.
• Chemistry: Chemical reaction rates and pH calculations.
• Computer Science: Algorithm analysis (logarithmic time complexity) and data compression.

Examples:

1. Compound Interest: If you invest $1000 at an annual interest rate of 5% compounded annually, the amount A after t years is given by the formula:

   • A = 1000(1 + 0.05)ᵗ

   To find out how long it takes for the investment to double, you would solve for t when A = 2000:

   • 2000 = 1000(1.05)ᵗ
   • 2 = (1.05)ᵗ
   • ln(2) = t * ln(1.05)
   • t = ln(2) / ln(1.05) ≈ 14.21 years

2. Radioactive Decay: The half-life of a radioactive substance is the time it takes for half of the substance to decay. If a substance has a half-life of 10 years, the amount N(t) remaining after t years is given by the formula:

   • N(t) = N₀(1/2)^(t/10), where N₀ is the initial amount.

   To find out how long it takes for 75% of the substance to decay (leaving 25% remaining), you would solve for t when N(t) = 0.25N₀:

   • 0.25N₀ = N₀(1/2)^(t/10)
   • 0.25 = (1/2)^(t/10)
   • log(0.25) = (t/10) * log(1/2)
   • t = 10 * (log(0.25) / log(0.5)) = 20 years

Graphing Exponential and Logarithmic Functions

Graphing these functions provides a visual representation of their behavior and properties.

Exponential Functions:

• f(x) = aˣ
  • If a > 1, the graph increases exponentially as x increases. The graph passes through (0, 1) and has a horizontal asymptote at y = 0.
  • If 0 < a < 1, the graph decreases exponentially as x increases. The graph passes through (0, 1) and has a horizontal asymptote at y = 0.

Logarithmic Functions:

• f(x) = logₐ(x)
  • If a > 1, the graph increases slowly as x increases. The graph passes through (1, 0) and has a vertical asymptote at x = 0.
  • If 0 < a < 1, the graph decreases slowly as x increases. The graph passes through (1, 0) and has a vertical asymptote at x = 0.

Transformations:

• Vertical Shift: f(x) + k shifts the graph up by k units if k > 0 and down by k units if k < 0.
• Horizontal Shift: f(x - h) shifts the graph right by h units if h > 0 and left by h units if h < 0.
• Vertical Stretch/Compression: af(x)* stretches the graph vertically by a factor of a if a > 1 and compresses it if 0 < a < 1.
• Horizontal Stretch/Compression: f(bx) compresses the graph horizontally by a factor of b if b > 1 and stretches it if 0 < b < 1.
• Reflection: -f(x) reflects the graph across the x-axis, and f(-x) reflects the graph across the y-axis.

Practice Problems: Mastering the Concepts

To solidify your understanding, practice solving various problems related to exponential and logarithmic functions.

Problem 1:

Solve for x: 4^(2x - 1) = 64

Solution:

• Rewrite 64 as a power of 4: 64 = 4³
• So, 4^(2x - 1) = 4³
• Equate the exponents: 2x - 1 = 3
• Solve for x: 2x = 4
• x = 2

Problem 2:

Solve for x: log₃(x) + log₃(x - 8) = 2

Solution:

• Combine the logarithms: log₃(x(x - 8)) = 2
• Exponentiate both sides: 3^(log₃(x(x - 8))) = 3²
• Simplify: x(x - 8) = 9
• Expand and solve the quadratic equation: x² - 8x = 9
• x² - 8x - 9 = 0
• (x - 9)(x + 1) = 0
• x = 9 or x = -1
• Check for extraneous solutions:
  • If x = -1, log₃(-1) is undefined. So, x = -1 is an extraneous solution.
  • If x = 9, log₃(9) + log₃(9 - 8) = log₃(9) + log₃(1) = 2 + 0 = 2. So, x = 9 is a valid solution.

Problem 3:

The population of a city is growing at a rate of 3% per year. If the current population is 50,000, what will the population be in 10 years?

Solution:

• Use the formula for exponential growth: P(t) = P₀(1 + r)ᵗ, where P(t) is the population after t years, P₀ is the initial population, and r is the growth rate.
• P(10) = 50000(1 + 0.03)¹⁰
• P(10) = 50000(1.03)¹⁰
• P(10) ≈ 67195.8
• The population will be approximately 67,196 in 10 years.

Problem 4:

A radioactive substance decays according to the formula N(t) = N₀e^(-kt), where N(t) is the amount remaining after t years, N₀ is the initial amount, and k is the decay constant. If the half-life of the substance is 50 years, find the value of k.

Solution:

• When t = 50, N(t) = 0.5N₀
• 0.5N₀ = N₀e^(-50k)
• 0.5 = e^(-50k)
• Take the natural logarithm of both sides: ln(0.5) = -50k
• Solve for k: k = ln(0.5) / -50
• k ≈ 0.01386

Common Mistakes to Avoid

• Forgetting to check for extraneous solutions when solving logarithmic equations.
• Incorrectly applying the laws of logarithms. Ensure you understand and apply the product, quotient, and power rules correctly.
• Confusing exponential growth and decay. Pay attention to the base a in the exponential function f(x) = aˣ. If a > 1, it's growth; if 0 < a < 1, it's decay.
• Misinterpreting the domain and range of exponential and logarithmic functions.
• Not understanding the inverse relationship between exponential and logarithmic functions.

Conclusion

Mastering Algebra 2 Unit 6, which focuses on exponential and logarithmic functions, is essential for building a strong foundation in mathematics. These functions are not just abstract concepts but powerful tools for modeling real-world phenomena. By understanding their properties, laws, and applications, and by practicing problem-solving, you can successfully navigate the challenges of this unit and excel in your mathematical studies. Remember to review the key concepts, practice consistently, and pay attention to common mistakes to solidify your understanding.