Learn on PengiOpenStax Algebra and TrigonometryChapter 6: Exponential and Logarithmic Functions

Lesson 6.1 : Exponential Functions

New Concept An exponential function models rapid growth or decay through a constant percent change. We'll master evaluating functions like $f(x) = ab^x$, finding their equations, and applying them to compound interest and continuous growth problems involving the number $e$.

Section 1

πŸ“˜ Exponential Functions

New Concept

An exponential function models rapid growth or decay through a constant percent change. We'll master evaluating functions like f(x)=abxf(x) = ab^x, finding their equations, and applying them to compound interest and continuous growth problems involving the number ee.

What’s next

This card is just the start. Next, you'll work through interactive examples of evaluating functions and tackle real-world challenge problems involving compound interest.

Section 2

Exponential Function

Property

For any real number xx, an exponential function is a function with the form

f(x)=abxf(x) = ab^x

where

  • aa is a non-zero real number called the initial value and
  • bb is any positive real number such that bβ‰ 1b \neq 1.
  • The domain of ff is all real numbers.
  • The range of ff is all positive real numbers if a>0a > 0.
  • The range of ff is all negative real numbers if a<0a < 0.
  • The yy-intercept is (0,a)(0, a), and the horizontal asymptote is y=0y = 0.

Section 3

Evaluating Exponential Functions

Property

To evaluate an exponential function with the form f(x)=bxf(x) = b^x, we simply substitute xx with the given value, and calculate the resulting power. For functions in the form f(x)=a(b)xf(x) = a(b)^x, it is important to follow the order of operations: simplify the power bxb^x first, then multiply by aa.

Examples

  • Let f(x)=4xf(x) = 4^x. To find f(3)f(3), substitute x=3x=3: f(3)=43=64f(3) = 4^3 = 64.
  • Let g(x)=10(2)x+1g(x) = 10(2)^{x+1}. To evaluate g(2)g(2), simplify the exponent first: g(2)=10(2)2+1=10(23)=10(8)=80g(2) = 10(2)^{2+1} = 10(2^3) = 10(8) = 80.
  • Let h(x)=9(13)xh(x) = 9(\frac{1}{3})^x. To find h(2)h(2), calculate the power first: h(2)=9(13)2=9(19)=1h(2) = 9(\frac{1}{3})^2 = 9(\frac{1}{9}) = 1.

Explanation

When evaluating, always handle the exponent first before any other operations like multiplication or addition. This ensures you're correctly calculating the exponential growth or decay before applying the initial value or other transformations.

Section 4

Exponential Growth

Property

A function that models exponential growth grows by a rate proportional to the amount present. For any real number xx and any positive real numbers aa and bb such that b≠1b \neq 1, an exponential growth function has the form

f(x)=abxf(x) = ab^x

where

  • aa is the initial or starting value of the function.
  • bb is the growth factor or growth multiplier per unit xx.

Section 5

Finding equation from two points

Property

To write an exponential model f(x)=abxf(x)=ab^x given two data points:

  1. If one of the data points has the form (0,a)(0, a), then aa is the initial value. Using aa, substitute the second point into the equation f(x)=abxf(x) = ab^x, and solve for bb.
  2. If neither of the data points have the form (0,a)(0, a), substitute both points into two equations with the form f(x)=abxf(x) = ab^x. Solve the resulting system of two equations in two unknowns to find aa and bb.

Examples

  • Given points (0,5)(0, 5) and (2,45)(2, 45): we know a=5a=5. Substitute the second point into y=5bxy=5b^x: 45=5b245 = 5b^2, so b2=9b^2=9, and b=3b=3. The equation is f(x)=5(3)xf(x)=5(3)^x.
  • Given points (2,12)(2, 12) and (4,108)(4, 108): create a system of equations: 12=ab212=ab^2 and 108=ab4108=ab^4. Divide the second by the first: 10812=ab4ab2\frac{108}{12} = \frac{ab^4}{ab^2}, which gives 9=b29=b^2, so b=3b=3. Substitute b=3b=3 into the first equation: 12=a(3)212 = a(3)^2, so a=129=43a=\frac{12}{9}=\frac{4}{3}. The equation is f(x)=43(3)xf(x)=\frac{4}{3}(3)^x.
  • A wolf population was 50 in 2010 and 80 in 2012. Letting t=0t=0 for 2010, the points are (0,50)(0, 50) and (2,80)(2, 80). Here, a=50a=50. Then 80=50b280=50b^2, so b=1.6β‰ˆ1.265b = \sqrt{1.6} \approx 1.265. The function is N(t)=50(1.265)tN(t)=50(1.265)^t.

Explanation

Two distinct points on a graph provide enough information to define a unique exponential curve. By substituting the coordinates into the general formula f(x)=abxf(x)=ab^x, you can solve for the initial value aa and the base bb.

Section 6

Compound Interest Formula

Property

Compound interest can be calculated using the formula

A(t)=P(1+rn)ntA(t) = P \left(1 + \frac{r}{n}\right)^{nt}

where

  • A(t)A(t) is the account value,
  • tt is measured in years,
  • PP is the starting amount of the account, often called the principal, or more generally present value,
  • rr is the annual percentage rate (APR) expressed as a decimal, and
  • nn is the number of compounding periods in one year.

Section 7

Continuous Growth and Base e

Property

The letter ee represents the irrational number

(1+1n)n,Β asΒ nΒ increasesΒ withoutΒ bound\left(1 + \frac{1}{n}\right)^n, \text{ as } n \text{ increases without bound}

Its approximation is eβ‰ˆ2.718282e \approx 2.718282. For all real numbers tt, and all positive numbers aa and rr, continuous growth or decay is represented by the formula

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Chapter 6: Exponential and Logarithmic Functions

  1. Lesson 1Current

    Lesson 6.1 : Exponential Functions

  2. Lesson 2

    Lesson 6.2 : Graphs of Exponential Functions

  3. Lesson 3

    Lesson 6.3 : Logarithmic Functions

  4. Lesson 4

    Lesson 6.4 : Graphs of Logarithmic Functions

  5. Lesson 5

    Lesson 6.5 : Logarithmic Properties

  6. Lesson 6

    Lesson 6.6 : Exponential and Logarithmic Equations

  7. Lesson 7

    Lesson 6.7 : Exponential and Logarithmic Models

  8. Lesson 8

    Lesson 6.8 : Fitting Exponential Models to Data

Lesson overview

Expand to review the lesson summary and core properties.

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Section 1

πŸ“˜ Exponential Functions

New Concept

An exponential function models rapid growth or decay through a constant percent change. We'll master evaluating functions like f(x)=abxf(x) = ab^x, finding their equations, and applying them to compound interest and continuous growth problems involving the number ee.

What’s next

This card is just the start. Next, you'll work through interactive examples of evaluating functions and tackle real-world challenge problems involving compound interest.

Section 2

Exponential Function

Property

For any real number xx, an exponential function is a function with the form

f(x)=abxf(x) = ab^x

where

  • aa is a non-zero real number called the initial value and
  • bb is any positive real number such that bβ‰ 1b \neq 1.
  • The domain of ff is all real numbers.
  • The range of ff is all positive real numbers if a>0a > 0.
  • The range of ff is all negative real numbers if a<0a < 0.
  • The yy-intercept is (0,a)(0, a), and the horizontal asymptote is y=0y = 0.

Section 3

Evaluating Exponential Functions

Property

To evaluate an exponential function with the form f(x)=bxf(x) = b^x, we simply substitute xx with the given value, and calculate the resulting power. For functions in the form f(x)=a(b)xf(x) = a(b)^x, it is important to follow the order of operations: simplify the power bxb^x first, then multiply by aa.

Examples

  • Let f(x)=4xf(x) = 4^x. To find f(3)f(3), substitute x=3x=3: f(3)=43=64f(3) = 4^3 = 64.
  • Let g(x)=10(2)x+1g(x) = 10(2)^{x+1}. To evaluate g(2)g(2), simplify the exponent first: g(2)=10(2)2+1=10(23)=10(8)=80g(2) = 10(2)^{2+1} = 10(2^3) = 10(8) = 80.
  • Let h(x)=9(13)xh(x) = 9(\frac{1}{3})^x. To find h(2)h(2), calculate the power first: h(2)=9(13)2=9(19)=1h(2) = 9(\frac{1}{3})^2 = 9(\frac{1}{9}) = 1.

Explanation

When evaluating, always handle the exponent first before any other operations like multiplication or addition. This ensures you're correctly calculating the exponential growth or decay before applying the initial value or other transformations.

Section 4

Exponential Growth

Property

A function that models exponential growth grows by a rate proportional to the amount present. For any real number xx and any positive real numbers aa and bb such that b≠1b \neq 1, an exponential growth function has the form

f(x)=abxf(x) = ab^x

where

  • aa is the initial or starting value of the function.
  • bb is the growth factor or growth multiplier per unit xx.

Section 5

Finding equation from two points

Property

To write an exponential model f(x)=abxf(x)=ab^x given two data points:

  1. If one of the data points has the form (0,a)(0, a), then aa is the initial value. Using aa, substitute the second point into the equation f(x)=abxf(x) = ab^x, and solve for bb.
  2. If neither of the data points have the form (0,a)(0, a), substitute both points into two equations with the form f(x)=abxf(x) = ab^x. Solve the resulting system of two equations in two unknowns to find aa and bb.

Examples

  • Given points (0,5)(0, 5) and (2,45)(2, 45): we know a=5a=5. Substitute the second point into y=5bxy=5b^x: 45=5b245 = 5b^2, so b2=9b^2=9, and b=3b=3. The equation is f(x)=5(3)xf(x)=5(3)^x.
  • Given points (2,12)(2, 12) and (4,108)(4, 108): create a system of equations: 12=ab212=ab^2 and 108=ab4108=ab^4. Divide the second by the first: 10812=ab4ab2\frac{108}{12} = \frac{ab^4}{ab^2}, which gives 9=b29=b^2, so b=3b=3. Substitute b=3b=3 into the first equation: 12=a(3)212 = a(3)^2, so a=129=43a=\frac{12}{9}=\frac{4}{3}. The equation is f(x)=43(3)xf(x)=\frac{4}{3}(3)^x.
  • A wolf population was 50 in 2010 and 80 in 2012. Letting t=0t=0 for 2010, the points are (0,50)(0, 50) and (2,80)(2, 80). Here, a=50a=50. Then 80=50b280=50b^2, so b=1.6β‰ˆ1.265b = \sqrt{1.6} \approx 1.265. The function is N(t)=50(1.265)tN(t)=50(1.265)^t.

Explanation

Two distinct points on a graph provide enough information to define a unique exponential curve. By substituting the coordinates into the general formula f(x)=abxf(x)=ab^x, you can solve for the initial value aa and the base bb.

Section 6

Compound Interest Formula

Property

Compound interest can be calculated using the formula

A(t)=P(1+rn)ntA(t) = P \left(1 + \frac{r}{n}\right)^{nt}

where

  • A(t)A(t) is the account value,
  • tt is measured in years,
  • PP is the starting amount of the account, often called the principal, or more generally present value,
  • rr is the annual percentage rate (APR) expressed as a decimal, and
  • nn is the number of compounding periods in one year.

Section 7

Continuous Growth and Base e

Property

The letter ee represents the irrational number

(1+1n)n,Β asΒ nΒ increasesΒ withoutΒ bound\left(1 + \frac{1}{n}\right)^n, \text{ as } n \text{ increases without bound}

Its approximation is eβ‰ˆ2.718282e \approx 2.718282. For all real numbers tt, and all positive numbers aa and rr, continuous growth or decay is represented by the formula

Book overview

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

Continue this chapter

Chapter 6: Exponential and Logarithmic Functions

  1. Lesson 1Current

    Lesson 6.1 : Exponential Functions

  2. Lesson 2

    Lesson 6.2 : Graphs of Exponential Functions

  3. Lesson 3

    Lesson 6.3 : Logarithmic Functions

  4. Lesson 4

    Lesson 6.4 : Graphs of Logarithmic Functions

  5. Lesson 5

    Lesson 6.5 : Logarithmic Properties

  6. Lesson 6

    Lesson 6.6 : Exponential and Logarithmic Equations

  7. Lesson 7

    Lesson 6.7 : Exponential and Logarithmic Models

  8. Lesson 8

    Lesson 6.8 : Fitting Exponential Models to Data