2.3 Exponential Functions Practice Answer Key

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Apr 19, 2025 · 7 min read

2.3 Exponential Functions Practice Answer Key
2.3 Exponential Functions Practice Answer Key

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    2.3 Exponential Functions Practice: Answers and In-Depth Explanations

    This comprehensive guide provides detailed answers and explanations for a typical set of practice problems on 2.3 Exponential Functions. We'll cover various aspects, from basic evaluations and graphing to more complex applications involving exponential growth and decay. Each problem will be tackled step-by-step, ensuring a thorough understanding of the underlying concepts. This resource aims to solidify your grasp of exponential functions and prepare you for more advanced topics.

    Understanding Exponential Functions

    Before diving into the practice problems, let's refresh our understanding of exponential functions. An exponential function is a function where the independent variable (usually x) appears as an exponent. The general form is:

    f(x) = a * b<sup>x</sup>

    Where:

    • a is the initial value (the y-intercept, where the graph crosses the y-axis).
    • b is the base, which determines the growth or decay rate.
      • If b > 1, the function represents exponential growth.
      • If 0 < b < 1, the function represents exponential decay.
      • If b = 1, the function is a constant function (f(x) = a).
      • If b ≤ 0, the function is not defined for all real numbers x.

    Key Concepts to Remember

    • Growth Factor: For exponential growth, the growth factor is (1 + r), where 'r' is the growth rate (often expressed as a percentage).
    • Decay Factor: For exponential decay, the decay factor is (1 - r), where 'r' is the decay rate.
    • Asymptotes: Exponential functions have a horizontal asymptote. For growth functions, it's the x-axis (y=0); for decay functions, it's also the x-axis.
    • Domain and Range: The domain of an exponential function is all real numbers (-∞, ∞). The range is (0, ∞) for both growth and decay functions (excluding the asymptote).

    Practice Problems and Solutions

    Let's tackle some practice problems. We'll categorize them for clarity.

    Section 1: Evaluating Exponential Functions

    Problem 1: Evaluate f(x) = 3<sup>x</sup> for x = 2, x = 0, and x = -1.

    Solution:

    • f(2) = 3<sup>2</sup> = 9
    • f(0) = 3<sup>0</sup> = 1 (Anything raised to the power of 0 is 1, except 0<sup>0</sup> which is undefined)
    • f(-1) = 3<sup>-1</sup> = 1/3 (A negative exponent means reciprocal)

    Problem 2: Evaluate g(x) = (1/2)<sup>x</sup> for x = 3, x = 0, and x = -2.

    Solution:

    • g(3) = (1/2)<sup>3</sup> = 1/8
    • g(0) = (1/2)<sup>0</sup> = 1
    • g(-2) = (1/2)<sup>-2</sup> = (2/1)<sup>2</sup> = 4

    Problem 3: If h(x) = 5 * 2<sup>x</sup>, find h(3) and h(-1).

    Solution:

    • h(3) = 5 * 2<sup>3</sup> = 5 * 8 = 40
    • h(-1) = 5 * 2<sup>-1</sup> = 5 * (1/2) = 5/2 = 2.5

    Section 2: Graphing Exponential Functions

    Problem 4: Graph the function f(x) = 2<sup>x</sup>. Identify the y-intercept and describe its behavior.

    Solution: Create a table of values:

    x f(x)
    -2 1/4
    -1 1/2
    0 1
    1 2
    2 4

    Plot these points. The y-intercept is (0, 1). The graph shows exponential growth; as x increases, f(x) increases rapidly. The x-axis acts as a horizontal asymptote.

    Problem 5: Graph the function g(x) = (1/3)<sup>x</sup>. Identify the y-intercept and describe its behavior.

    Solution: Similar to Problem 4, create a table of values:

    x g(x)
    -2 9
    -1 3
    0 1
    1 1/3
    2 1/9

    Plot these points. The y-intercept is (0,1). This graph exhibits exponential decay; as x increases, g(x) approaches 0. The x-axis is the horizontal asymptote.

    Section 3: Applications of Exponential Growth and Decay

    Problem 6: A population of bacteria doubles every hour. If the initial population is 1000, what will the population be after 3 hours?

    Solution: This is exponential growth. The formula is: P(t) = P<sub>0</sub> * 2<sup>t</sup>, where P<sub>0</sub> is the initial population and t is the time in hours.

    P(3) = 1000 * 2<sup>3</sup> = 1000 * 8 = 8000

    The population will be 8000 after 3 hours.

    Problem 7: The value of a car depreciates at a rate of 15% per year. If the initial value is $20,000, what will its value be after 2 years?

    Solution: This is exponential decay. The decay factor is (1 - 0.15) = 0.85. The formula is: V(t) = V<sub>0</sub> * (0.85)<sup>t</sup>, where V<sub>0</sub> is the initial value and t is the time in years.

    V(2) = 20000 * (0.85)<sup>2</sup> = 20000 * 0.7225 = $14450

    The car's value will be $14,450 after 2 years.

    Problem 8: A radioactive substance decays according to the equation A(t) = A<sub>0</sub> * e<sup>-kt</sup>, where A<sub>0</sub> is the initial amount, k is the decay constant, and t is the time. If A<sub>0</sub> = 100 grams and k = 0.05, find the amount remaining after 10 years. (Note: 'e' is Euler's number, approximately 2.718)

    Solution:

    A(10) = 100 * e<sup>-0.05*10</sup> = 100 * e<sup>-0.5</sup>

    Using a calculator to find e<sup>-0.5</sup> ≈ 0.6065

    A(10) ≈ 100 * 0.6065 = 60.65 grams

    Approximately 60.65 grams will remain after 10 years.

    Section 4: Solving Exponential Equations

    Problem 9: Solve for x: 2<sup>x</sup> = 8

    Solution: We can rewrite 8 as 2<sup>3</sup>. Therefore:

    2<sup>x</sup> = 2<sup>3</sup>

    This implies x = 3

    Problem 10: Solve for x: 3<sup>x</sup> = 1/9

    Solution: We can rewrite 1/9 as 3<sup>-2</sup>. Therefore:

    3<sup>x</sup> = 3<sup>-2</sup>

    This implies x = -2

    Problem 11: Solve for x: 5<sup>x+1</sup> = 25

    Solution: Rewrite 25 as 5<sup>2</sup>:

    5<sup>x+1</sup> = 5<sup>2</sup>

    Therefore, x + 1 = 2, which means x = 1

    Section 5: More Complex Problems

    Problem 12: The number of subscribers to a social media platform increases by 20% each month. If there are currently 10,000 subscribers, how many will there be in 6 months?

    Solution: This is exponential growth. The growth factor is 1.2 (1 + 0.2). The formula is: S(t) = S<sub>0</sub> * (1.2)<sup>t</sup>, where S<sub>0</sub> is the initial number of subscribers and t is the time in months.

    S(6) = 10000 * (1.2)<sup>6</sup> ≈ 29859.84

    Rounding to the nearest whole number, there will be approximately 29,860 subscribers in 6 months.

    Problem 13: A certain medication has a half-life of 4 hours. If a patient takes a 200mg dose, how much will remain in their system after 12 hours?

    Solution: This is exponential decay. The decay factor is 0.5 (1/2). The formula is: M(t) = M<sub>0</sub> * (0.5)<sup>t/h</sup>, where M<sub>0</sub> is the initial dosage, t is the time elapsed, and h is the half-life.

    M(12) = 200 * (0.5)<sup>12/4</sup> = 200 * (0.5)<sup>3</sup> = 200 * (1/8) = 25 mg

    25mg of the medication will remain after 12 hours.

    This comprehensive guide provides a detailed walkthrough of various problems involving exponential functions. Remember to practice regularly to master these concepts. Understanding the underlying principles, formulas, and the different types of problems will greatly enhance your ability to solve exponential function questions effectively. Remember to utilize online resources and textbooks to further enhance your understanding and practice more complex problems. Consistent practice will build your confidence and proficiency in this crucial mathematical area.

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