2.3 Exponential Functions Practice Answer Key

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May 25, 2025 · 6 min read

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2.3 Exponential Functions Practice: A Comprehensive Guide with Answer Key
Exponential functions are a cornerstone of mathematics, appearing in diverse fields from finance and biology to physics and computer science. Understanding them is crucial for success in many academic pursuits and professional careers. This article provides a comprehensive guide to solving problems related to 2.3 exponential functions, complete with an answer key and explanations to help you master this topic. We'll cover various types of problems, offering detailed solutions and strategies to enhance your problem-solving skills. This in-depth exploration ensures a thorough understanding of exponential functions, improving your ability to apply these concepts effectively.
Understanding Exponential Functions
Before diving into practice problems, let's solidify our understanding of exponential functions. An exponential function is a function of the form:
f(x) = ab<sup>x</sup>
where:
- a is the initial value (or y-intercept, the value of the function when x = 0).
- b is the base, a constant representing the rate of growth or decay. If b > 1, the function represents exponential growth; if 0 < b < 1, it represents exponential decay.
- x is the exponent, often representing time or some other independent variable.
Understanding the roles of 'a' and 'b' is fundamental to solving exponential function problems. 'a' determines the starting point of the function on the y-axis, while 'b' dictates how quickly the function increases or decreases.
Key Properties of Exponential Functions
- Domain: All real numbers (-∞, ∞)
- Range: (0, ∞) if a > 0 and b > 0 (assuming the function is strictly exponential)
- Asymptote: The x-axis (y = 0) is a horizontal asymptote for exponential functions.
- One-to-one function: Each x-value corresponds to a unique y-value, and vice-versa. This allows for the use of inverse functions (logarithms).
- Growth vs Decay: If b > 1, it's exponential growth; if 0 < b < 1, it's exponential decay.
Types of Problems in 2.3 Exponential Functions Practice
Practice problems for exponential functions cover a wide range of concepts, including:
1. Evaluating Exponential Functions
This involves substituting a given value of 'x' into the equation f(x) = ab<sup>x</sup> to find the corresponding y-value.
Example: If f(x) = 2(3)<sup>x</sup>, find f(2).
Solution: f(2) = 2(3)<sup>2</sup> = 2(9) = 18
2. Finding the Equation of an Exponential Function
Given points on the graph of an exponential function, we can determine the equation. This often involves solving a system of equations.
Example: Find the equation of an exponential function that passes through points (1, 6) and (2, 18).
Solution: Let the equation be f(x) = ab<sup>x</sup>. Substituting the points gives us:
6 = ab<sup>1</sup> 18 = ab<sup>2</sup>
Dividing the second equation by the first, we get:
3 = b
Substituting b = 3 into the first equation:
6 = a(3) => a = 2
Therefore, the equation is f(x) = 2(3)<sup>x</sup>
3. Solving Exponential Equations
This involves finding the value of x that satisfies an exponential equation. Often, this requires using logarithms.
Example: Solve 2<sup>x</sup> = 16
Solution: Since 16 = 2<sup>4</sup>, we have 2<sup>x</sup> = 2<sup>4</sup>. Therefore, x = 4.
4. Applications of Exponential Functions
Exponential functions model many real-world phenomena, including population growth, radioactive decay, and compound interest. Problems in this area often involve applying the exponential function formula to a specific context.
Example: A population of bacteria doubles every hour. If the initial population is 1000, what is the population after 3 hours?
Solution: The equation is P(t) = 1000(2)<sup>t</sup>, where t is the time in hours. After 3 hours, the population is P(3) = 1000(2)<sup>3</sup> = 8000.
5. Graphing Exponential Functions
Understanding how to graph exponential functions is crucial for visualizing their behavior and identifying key features like asymptotes and intercepts.
Example: Graph the function f(x) = 2<sup>x</sup>
Solution: Create a table of values, plot the points, and connect them with a smooth curve. Remember that the graph approaches but never touches the x-axis (the asymptote).
2.3 Exponential Functions Practice Problems with Answer Key
Now let's tackle some practice problems, categorized by difficulty level. Remember to show your work! The answer key is provided at the end.
Beginner:
- If f(x) = 5(2)<sup>x</sup>, find f(3).
- If g(x) = (1/2)<sup>x</sup>, find g(2) and g(-1).
- What is the y-intercept of the function h(x) = 3(4)<sup>x</sup>?
Intermediate:
- Find the equation of an exponential function that passes through points (0, 2) and (1, 6).
- Solve the equation 3<sup>x</sup> = 81.
- A certain investment grows by 5% each year. If the initial investment is $1000, how much will it be worth after 2 years? (Round to the nearest cent)
Advanced:
- The half-life of a radioactive substance is 10 years. If the initial amount is 100 grams, how much will remain after 20 years?
- Solve the equation 2<sup>2x</sup> = 8<sup>x-1</sup>
- Find the equation of an exponential function that passes through points (1, 4) and (3, 16). What is its y-intercept?
Answer Key and Explanations
Beginner:
- f(3) = 5(2)<sup>3</sup> = 40
- g(2) = (1/2)<sup>2</sup> = 1/4; g(-1) = (1/2)<sup>-1</sup> = 2
- The y-intercept is the value when x = 0. h(0) = 3(4)<sup>0</sup> = 3(1) = 3
Intermediate:
- Let f(x) = ab<sup>x</sup>. Using (0,2) and (1,6): 2 = a(b)<sup>0</sup> => a = 2; 6 = 2b<sup>1</sup> => b = 3. Therefore, f(x) = 2(3)<sup>x</sup>.
- 3<sup>x</sup> = 81 = 3<sup>4</sup>. Therefore, x = 4.
- A = P(1 + r)<sup>t</sup>, where A is the final amount, P is the principal, r is the interest rate, and t is the time. A = 1000(1 + 0.05)<sup>2</sup> = 1000(1.05)<sup>2</sup> ≈ $1102.50
Advanced:
- Let A(t) = A<sub>0</sub>(1/2)<sup>t/h</sup>, where A<sub>0</sub> is the initial amount, t is the time, and h is the half-life. A(20) = 100(1/2)<sup>20/10</sup> = 100(1/2)<sup>2</sup> = 25 grams.
- 2<sup>2x</sup> = 8<sup>x-1</sup> can be written as 2<sup>2x</sup> = (2<sup>3</sup>)<sup>x-1</sup> = 2<sup>3x-3</sup>. Therefore, 2x = 3x - 3, which gives x = 3.
- Let f(x) = ab<sup>x</sup>. Using (1,4) and (3,16): 4 = ab; 16 = ab<sup>3</sup>. Dividing the second by the first: 4 = b<sup>2</sup> => b = 2 (since b must be positive). Substituting b = 2 into 4 = ab: a = 2. Therefore, f(x) = 2(2)<sup>x</sup> = 2<sup>x+1</sup>. The y-intercept is when x = 0: f(0) = 2<sup>1</sup> = 2.
This comprehensive guide and the accompanying answer key provide a strong foundation for mastering 2.3 exponential functions. Remember that consistent practice and a clear understanding of the underlying concepts are key to success. Through dedicated work on problems of varying difficulty, you will develop a solid grasp of exponential functions and their applications. This knowledge will be invaluable in various fields of study and future endeavors.
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