Learn on PengiOpenstax Intermediate Algebra 2EChapter 10: Exponential and Logarithmic Functions

Lesson 10.2: Evaluate and Graph Exponential Functions

In this lesson from OpenStax Intermediate Algebra 2E, students learn to define and graph exponential functions of the form f(x) = a^x, exploring key properties such as domain, range, horizontal asymptotes, and characteristic points like (0, 1) and (1, a). Learners compare exponential growth curves for bases greater than 1, solve exponential equations, and apply these models to real-world contexts such as population growth, radioactive decay, and compound interest.

Section 1

πŸ“˜ Evaluate and Graph Exponential Functions

New Concept

Explore exponential functions, where the variable is in the exponent, like in f(x)=axf(x) = a^x. You'll learn to graph their distinct shapes, solve equations, and apply these powerful models to real-world scenarios like compound interest and population growth.

What’s next

Next, you'll tackle practice cards that break down graphing and solving. Then, you'll apply these skills to challenge problems on compound interest.

Section 2

Exponential Function

Property

An exponential function, where a>0a > 0 and a≠1a \neq 1, is a function of the form

f(x)=axf(x) = a^x

Examples

  • The function f(x)=5xf(x) = 5^x is an exponential function because the variable xx is the exponent and the base 55 is a positive number not equal to 11.
  • The function g(x)=x4g(x) = x^4 is a polynomial, not an exponential function. Here, the variable xx is the base, not the exponent.
  • A function like h(x)=(βˆ’3)xh(x) = (-3)^x is not a valid exponential function because its base is negative, which results in non-real numbers for fractional exponents like x=12x = \frac{1}{2}.

Explanation

Unlike functions where the variable xx is the base, here the variable is the exponent! This structure causes the function's value to grow or shrink incredibly fast, making it perfect for modeling real-world phenomena like population growth.

Section 3

Properties of Exponential Graphs

Property

Properties of the Graph of f(x)=axf(x) = a^x

when a>1a > 1when 0<a<10 < a < 1
Domain(βˆ’βˆž,∞)(-\infty, \infty)(βˆ’βˆž,∞)(-\infty, \infty)
Range(0,∞)(0, \infty)(0,∞)(0, \infty)
x-interceptnonenone
y-intercept(0,1)(0, 1)(0,1)(0, 1)
Contains(1,a),(βˆ’1,1a)(1, a), (-1, \frac{1}{a})(1,a),(βˆ’1,1a)(1, a), (-1, \frac{1}{a})
Asymptotex-axis, the line y=0y=0x-axis, the line y=0y=0
Basic shapeincreasingdecreasing

Examples

  • The graph of f(x)=4xf(x) = 4^x is an increasing curve because a>1a > 1. It passes through the points (0,1)(0, 1) and (1,4)(1, 4).
  • The graph of g(x)=(14)xg(x) = (\frac{1}{4})^x is a decreasing curve because 0<a<10 < a < 1. It passes through the points (0,1)(0, 1) and (βˆ’1,4)(-1, 4).
  • Both graphs f(x)=4xf(x) = 4^x and g(x)=(14)xg(x) = (\frac{1}{4})^x have a domain of all real numbers, (βˆ’βˆž,∞)(-\infty, \infty), and a range of all positive numbers, (0,∞)(0, \infty).

Section 4

Graphing with Translations

Property

For an exponential function f(x)=axf(x)=a^x, the graph can be translated:

  1. Horizontal shift: The graph of g(x)=axβˆ’hg(x) = a^{x-h} is the graph of f(x)f(x) shifted hh units horizontally.
  2. Vertical shift: The graph of g(x)=ax+kg(x) = a^x + k is the graph of f(x)f(x) shifted kk units vertically. The horizontal asymptote also shifts to y=ky=k.

Examples

  • The graph of g(x)=2xβˆ’3g(x) = 2^{x-3} is the graph of f(x)=2xf(x) = 2^x shifted 3 units to the right. The point (0,1)(0, 1) on f(x)f(x) moves to (3,1)(3, 1) on g(x)g(x).
  • The graph of h(x)=3x+5h(x) = 3^x + 5 is the graph of f(x)=3xf(x) = 3^x shifted 5 units up. The horizontal asymptote moves from y=0y=0 to y=5y=5.
  • To graph k(x)=4x+1βˆ’2k(x) = 4^{x+1} - 2, take the graph of f(x)=4xf(x)=4^x, shift it 1 unit to the left, and then 2 units down.

Explanation

Think of it as moving the entire picture of the graph. Adding or subtracting inside the exponent slides the graph left or right. Adding or subtracting outside the function moves it up or down, taking the horizontal asymptote with it.

Section 5

Natural Base e

Property

Natural Base e: The number ee is defined as the value of (1+1n)n(1 + \frac{1}{n})^n, as nn increases without bound. We say, as nn approaches infinity,
eβ‰ˆ2.718281827…e \approx 2.718281827\ldots

Natural Exponential Function: The natural exponential function is an exponential function whose base is ee

f(x)=exf(x) = e^x

The domain is (βˆ’βˆž,∞)(-\infty, \infty) and the range is (0,∞)(0, \infty).

Examples

  • The graph of the natural exponential function f(x)=exf(x) = e^x is an increasing curve that sits between the graphs of g(x)=2xg(x) = 2^x and h(x)=3xh(x) = 3^x, since 2<e<32 < e < 3.
  • As you calculate (1+1n)n(1 + \frac{1}{n})^n for larger and larger values of nn, the result gets closer and closer to eβ‰ˆ2.718281827e \approx 2.718281827.
  • Using a calculator, the value of e2e^2 is approximately 7.3897.389. This corresponds to the point (2,7.389)(2, 7.389) on the graph of f(x)=exf(x)=e^x.

Section 6

One-to-One Property

Property

One-to-One Property of Exponential Equations
For a>0a > 0 and a≠1a \neq 1,

If⁑ ax=ay, then⁑ x=y. \operatorname{If}\ a^x = a^y,\ \operatorname{then}\ x = y.

To solve an exponential equation:

  1. Write both sides of the equation with the same base, if possible.
  2. Write a new equation by setting the exponents equal.
  3. Solve the new equation.
  4. Check the solution.

Examples

  • To solve 4x+2=644^{x+2} = 64, first rewrite 6464 as 434^3. The equation becomes 4x+2=434^{x+2} = 4^3. Now, set the exponents equal: x+2=3x+2 = 3, which gives x=1x=1.
  • Solve ex2βˆ’3=e2xe^{x^2-3} = e^{2x}. Since the bases are both ee, we set the exponents equal: x2βˆ’3=2xx^2-3 = 2x. This gives a quadratic equation x2βˆ’2xβˆ’3=0x^2-2x-3=0, which factors to (xβˆ’3)(x+1)=0(x-3)(x+1)=0. So, x=3x=3 or x=βˆ’1x=-1.
  • Solve 9x=279^{x} = 27. Write both sides with base 3: (32)x=33(3^2)^x = 3^3, which simplifies to 32x=333^{2x} = 3^3. Therefore, 2x=32x=3, and x=32x=\frac{3}{2}.

Explanation

This property is a powerful shortcut for solving exponential equations. If you can make the bases on both sides of the equation the same, you can ignore the bases and simply set the exponents equal to each other to solve.

Section 7

Compound Interest

Property

For a principal, PP, invested at an interest rate, rr, for tt years, the new balance, AA, is:

A=P(1+rn)ntwhenΒ compoundedΒ nΒ timesΒ aΒ year. A = P \left(1 + \frac{r}{n}\right)^{nt} \quad \text{when compounded n times a year.}
A=PertwhenΒ compoundedΒ continuously. A = Pe^{rt} \quad \text{when compounded continuously.}

Examples

  • To find the amount after 10 years on a 5,000 dollars investment at 4%4\% interest compounded quarterly, use A=5000(1+0.044)4β‹…10A = 5000(1 + \frac{0.04}{4})^{4 \cdot 10}.
  • For the same 5,000 dollars at 4%4\% for 10 years compounded monthly, the formula is A=5000(1+0.0412)12β‹…10A = 5000(1 + \frac{0.04}{12})^{12 \cdot 10}.
  • If the 5,000 dollars is invested at 4%4\% for 10 years with interest compounded continuously, you use the formula A=5000e0.04β‹…10A = 5000e^{0.04 \cdot 10}.

Explanation

This is how money grows over time by earning interest on itself. The first formula is for interest calculated at set intervals (like monthly). The second, using base 'e', is for interest that is calculated and added constantly.

Section 8

Exponential Growth and Decay

Property

For an original amount, A0A_0, that grows or decays at a rate, rr, for a certain time, tt, the final amount, AA, is:

A=A0ert A = A_0 e^{rt}

Examples

  • A population of 200 rabbits grows continuously at a rate of 15%15\% per year. After 5 years, the population will be A=200e0.15β‹…5A = 200e^{0.15 \cdot 5}.
  • A chemical has a half-life that corresponds to a continuous decay rate of r=βˆ’0.05r = -0.05 per hour. If you start with 100 grams, the amount left after 24 hours is A=100eβˆ’0.05β‹…24A = 100e^{-0.05 \cdot 24}.
  • A city's population of 500,000 is growing continuously at a rate of 2%2\% per year. The population in tt years can be modeled by the equation A=500000e0.02tA = 500000e^{0.02t}.

Explanation

This formula models anything that changes by a constant percentage over time. A positive rate 'r' means growth, like a population of bacteria. A negative rate 'r' signifies decay, like a radioactive substance losing mass over time.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 10: Exponential and Logarithmic Functions

  1. Lesson 1

    Lesson 10.1: Finding Composite and Inverse Functions

  2. Lesson 2Current

    Lesson 10.2: Evaluate and Graph Exponential Functions

  3. Lesson 3

    Lesson 10.3: Evaluate and Graph Logarithmic Functions

  4. Lesson 4

    Lesson 10.4: Use the Properties of Logarithms

  5. Lesson 5

    Lesson 10.5: Solve Exponential and Logarithmic Equations

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Evaluate and Graph Exponential Functions

New Concept

Explore exponential functions, where the variable is in the exponent, like in f(x)=axf(x) = a^x. You'll learn to graph their distinct shapes, solve equations, and apply these powerful models to real-world scenarios like compound interest and population growth.

What’s next

Next, you'll tackle practice cards that break down graphing and solving. Then, you'll apply these skills to challenge problems on compound interest.

Section 2

Exponential Function

Property

An exponential function, where a>0a > 0 and a≠1a \neq 1, is a function of the form

f(x)=axf(x) = a^x

Examples

  • The function f(x)=5xf(x) = 5^x is an exponential function because the variable xx is the exponent and the base 55 is a positive number not equal to 11.
  • The function g(x)=x4g(x) = x^4 is a polynomial, not an exponential function. Here, the variable xx is the base, not the exponent.
  • A function like h(x)=(βˆ’3)xh(x) = (-3)^x is not a valid exponential function because its base is negative, which results in non-real numbers for fractional exponents like x=12x = \frac{1}{2}.

Explanation

Unlike functions where the variable xx is the base, here the variable is the exponent! This structure causes the function's value to grow or shrink incredibly fast, making it perfect for modeling real-world phenomena like population growth.

Section 3

Properties of Exponential Graphs

Property

Properties of the Graph of f(x)=axf(x) = a^x

when a>1a > 1when 0<a<10 < a < 1
Domain(βˆ’βˆž,∞)(-\infty, \infty)(βˆ’βˆž,∞)(-\infty, \infty)
Range(0,∞)(0, \infty)(0,∞)(0, \infty)
x-interceptnonenone
y-intercept(0,1)(0, 1)(0,1)(0, 1)
Contains(1,a),(βˆ’1,1a)(1, a), (-1, \frac{1}{a})(1,a),(βˆ’1,1a)(1, a), (-1, \frac{1}{a})
Asymptotex-axis, the line y=0y=0x-axis, the line y=0y=0
Basic shapeincreasingdecreasing

Examples

  • The graph of f(x)=4xf(x) = 4^x is an increasing curve because a>1a > 1. It passes through the points (0,1)(0, 1) and (1,4)(1, 4).
  • The graph of g(x)=(14)xg(x) = (\frac{1}{4})^x is a decreasing curve because 0<a<10 < a < 1. It passes through the points (0,1)(0, 1) and (βˆ’1,4)(-1, 4).
  • Both graphs f(x)=4xf(x) = 4^x and g(x)=(14)xg(x) = (\frac{1}{4})^x have a domain of all real numbers, (βˆ’βˆž,∞)(-\infty, \infty), and a range of all positive numbers, (0,∞)(0, \infty).

Section 4

Graphing with Translations

Property

For an exponential function f(x)=axf(x)=a^x, the graph can be translated:

  1. Horizontal shift: The graph of g(x)=axβˆ’hg(x) = a^{x-h} is the graph of f(x)f(x) shifted hh units horizontally.
  2. Vertical shift: The graph of g(x)=ax+kg(x) = a^x + k is the graph of f(x)f(x) shifted kk units vertically. The horizontal asymptote also shifts to y=ky=k.

Examples

  • The graph of g(x)=2xβˆ’3g(x) = 2^{x-3} is the graph of f(x)=2xf(x) = 2^x shifted 3 units to the right. The point (0,1)(0, 1) on f(x)f(x) moves to (3,1)(3, 1) on g(x)g(x).
  • The graph of h(x)=3x+5h(x) = 3^x + 5 is the graph of f(x)=3xf(x) = 3^x shifted 5 units up. The horizontal asymptote moves from y=0y=0 to y=5y=5.
  • To graph k(x)=4x+1βˆ’2k(x) = 4^{x+1} - 2, take the graph of f(x)=4xf(x)=4^x, shift it 1 unit to the left, and then 2 units down.

Explanation

Think of it as moving the entire picture of the graph. Adding or subtracting inside the exponent slides the graph left or right. Adding or subtracting outside the function moves it up or down, taking the horizontal asymptote with it.

Section 5

Natural Base e

Property

Natural Base e: The number ee is defined as the value of (1+1n)n(1 + \frac{1}{n})^n, as nn increases without bound. We say, as nn approaches infinity,
eβ‰ˆ2.718281827…e \approx 2.718281827\ldots

Natural Exponential Function: The natural exponential function is an exponential function whose base is ee

f(x)=exf(x) = e^x

The domain is (βˆ’βˆž,∞)(-\infty, \infty) and the range is (0,∞)(0, \infty).

Examples

  • The graph of the natural exponential function f(x)=exf(x) = e^x is an increasing curve that sits between the graphs of g(x)=2xg(x) = 2^x and h(x)=3xh(x) = 3^x, since 2<e<32 < e < 3.
  • As you calculate (1+1n)n(1 + \frac{1}{n})^n for larger and larger values of nn, the result gets closer and closer to eβ‰ˆ2.718281827e \approx 2.718281827.
  • Using a calculator, the value of e2e^2 is approximately 7.3897.389. This corresponds to the point (2,7.389)(2, 7.389) on the graph of f(x)=exf(x)=e^x.

Section 6

One-to-One Property

Property

One-to-One Property of Exponential Equations
For a>0a > 0 and a≠1a \neq 1,

If⁑ ax=ay, then⁑ x=y. \operatorname{If}\ a^x = a^y,\ \operatorname{then}\ x = y.

To solve an exponential equation:

  1. Write both sides of the equation with the same base, if possible.
  2. Write a new equation by setting the exponents equal.
  3. Solve the new equation.
  4. Check the solution.

Examples

  • To solve 4x+2=644^{x+2} = 64, first rewrite 6464 as 434^3. The equation becomes 4x+2=434^{x+2} = 4^3. Now, set the exponents equal: x+2=3x+2 = 3, which gives x=1x=1.
  • Solve ex2βˆ’3=e2xe^{x^2-3} = e^{2x}. Since the bases are both ee, we set the exponents equal: x2βˆ’3=2xx^2-3 = 2x. This gives a quadratic equation x2βˆ’2xβˆ’3=0x^2-2x-3=0, which factors to (xβˆ’3)(x+1)=0(x-3)(x+1)=0. So, x=3x=3 or x=βˆ’1x=-1.
  • Solve 9x=279^{x} = 27. Write both sides with base 3: (32)x=33(3^2)^x = 3^3, which simplifies to 32x=333^{2x} = 3^3. Therefore, 2x=32x=3, and x=32x=\frac{3}{2}.

Explanation

This property is a powerful shortcut for solving exponential equations. If you can make the bases on both sides of the equation the same, you can ignore the bases and simply set the exponents equal to each other to solve.

Section 7

Compound Interest

Property

For a principal, PP, invested at an interest rate, rr, for tt years, the new balance, AA, is:

A=P(1+rn)ntwhenΒ compoundedΒ nΒ timesΒ aΒ year. A = P \left(1 + \frac{r}{n}\right)^{nt} \quad \text{when compounded n times a year.}
A=PertwhenΒ compoundedΒ continuously. A = Pe^{rt} \quad \text{when compounded continuously.}

Examples

  • To find the amount after 10 years on a 5,000 dollars investment at 4%4\% interest compounded quarterly, use A=5000(1+0.044)4β‹…10A = 5000(1 + \frac{0.04}{4})^{4 \cdot 10}.
  • For the same 5,000 dollars at 4%4\% for 10 years compounded monthly, the formula is A=5000(1+0.0412)12β‹…10A = 5000(1 + \frac{0.04}{12})^{12 \cdot 10}.
  • If the 5,000 dollars is invested at 4%4\% for 10 years with interest compounded continuously, you use the formula A=5000e0.04β‹…10A = 5000e^{0.04 \cdot 10}.

Explanation

This is how money grows over time by earning interest on itself. The first formula is for interest calculated at set intervals (like monthly). The second, using base 'e', is for interest that is calculated and added constantly.

Section 8

Exponential Growth and Decay

Property

For an original amount, A0A_0, that grows or decays at a rate, rr, for a certain time, tt, the final amount, AA, is:

A=A0ert A = A_0 e^{rt}

Examples

  • A population of 200 rabbits grows continuously at a rate of 15%15\% per year. After 5 years, the population will be A=200e0.15β‹…5A = 200e^{0.15 \cdot 5}.
  • A chemical has a half-life that corresponds to a continuous decay rate of r=βˆ’0.05r = -0.05 per hour. If you start with 100 grams, the amount left after 24 hours is A=100eβˆ’0.05β‹…24A = 100e^{-0.05 \cdot 24}.
  • A city's population of 500,000 is growing continuously at a rate of 2%2\% per year. The population in tt years can be modeled by the equation A=500000e0.02tA = 500000e^{0.02t}.

Explanation

This formula models anything that changes by a constant percentage over time. A positive rate 'r' means growth, like a population of bacteria. A negative rate 'r' signifies decay, like a radioactive substance losing mass over time.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 10: Exponential and Logarithmic Functions

  1. Lesson 1

    Lesson 10.1: Finding Composite and Inverse Functions

  2. Lesson 2Current

    Lesson 10.2: Evaluate and Graph Exponential Functions

  3. Lesson 3

    Lesson 10.3: Evaluate and Graph Logarithmic Functions

  4. Lesson 4

    Lesson 10.4: Use the Properties of Logarithms

  5. Lesson 5

    Lesson 10.5: Solve Exponential and Logarithmic Equations