Lesson Objectives

- Demonstrate an understanding of the exponential function: f(x) = a
^{x} - Learn about the logarithmic function: f(x) = log
_{a}x - Learn how to convert between logarithmic form and exponential form
- Learn how to solve simple logarithmic equations
- Learn how to graph a logarithmic function

## How to Work with Logarithmic Functions

We previously learned about the exponential function.

f(x) = a

A logarithmic function is the inverse of the exponential function.

f(x) = log

We can derive this function by following the steps needed to find the inverse of a function. Let's begin with the exponential function.

f(x) = a

Let's replace f(x) with y:

y = a

Now, we can swap x and y:

x = a

At this point, we want to solve the equation for y. How can we do this? Up to this point, we have not seen any method that allows us to solve an equation for the exponent. This is why we introduce the logarithm. A logarithm is an exponent. We will set our exponent of y equal to the base a logarithm of x:

y = log

log

It is very important to memorize these two generic forms. This will allow us to go back and forth between exponential and logarithmic forms. Let's look at a few examples.

Example 1: Convert each to logarithmic form. $$17^2=289$$ To convert this into logarithmic form, follow the model above. $$2=\log_{17}(289)$$ Again, this form allows us to isolate the exponent. We are saying that log base 17 of 289 is 2. This just means that 2 is the exponent required on 17 to obtain the number 289.

Example 2: Convert each to logarithmic form. $$7^{-2}=\frac{1}{49}$$ To convert this into logarithmic form, follow the model above. $$-2=\log_{7}\left(\frac{1}{49}\right)$$ Again, this form allows us to isolate the exponent. We are saying that log base 7 of 1/49 is -2. This just means that -2 is the exponent required on 7 to obtain the number 1/49.

Example 3: Convert each to exponential form. $$-\frac{1}{2}=\log_{64}\left(\frac{1}{8}\right)$$ To convert this into exponential form, follow the model above. $$64^{-\frac{1}{2}}=\frac{1}{8}$$

f(x) = a

^{x}, a > 0, a ≠ 1A logarithmic function is the inverse of the exponential function.

f(x) = log

_{a}x, a > 0, a ≠ 1, x > 0We can derive this function by following the steps needed to find the inverse of a function. Let's begin with the exponential function.

f(x) = a

^{x}, a > 0, a ≠ 1Let's replace f(x) with y:

y = a

^{x}Now, we can swap x and y:

x = a

^{y}At this point, we want to solve the equation for y. How can we do this? Up to this point, we have not seen any method that allows us to solve an equation for the exponent. This is why we introduce the logarithm. A logarithm is an exponent. We will set our exponent of y equal to the base a logarithm of x:

y = log

_{a}xlog

_{a}x is the exponent to which the base a must be raised to obtain x. This means we have two forms for the same thing:Log Form: | Exponential Form: |
---|---|

y = log_{a}x | x = a^{y} |

Example 1: Convert each to logarithmic form. $$17^2=289$$ To convert this into logarithmic form, follow the model above. $$2=\log_{17}(289)$$ Again, this form allows us to isolate the exponent. We are saying that log base 17 of 289 is 2. This just means that 2 is the exponent required on 17 to obtain the number 289.

Example 2: Convert each to logarithmic form. $$7^{-2}=\frac{1}{49}$$ To convert this into logarithmic form, follow the model above. $$-2=\log_{7}\left(\frac{1}{49}\right)$$ Again, this form allows us to isolate the exponent. We are saying that log base 7 of 1/49 is -2. This just means that -2 is the exponent required on 7 to obtain the number 1/49.

Example 3: Convert each to exponential form. $$-\frac{1}{2}=\log_{64}\left(\frac{1}{8}\right)$$ To convert this into exponential form, follow the model above. $$64^{-\frac{1}{2}}=\frac{1}{8}$$

## Solving Logarithmic Equations

We will cover how to solve more advanced logarithmic equations in a future lesson. For this lesson, let's talk about how to solve a simple logarithmic equation of the form: $$\log_{a}(x)=k$$ To solve this type of equation, we place the equation in exponential form. Let's look at a few examples.

Example 4: Solve each equation. $$-2\log_{3}(5x + 9) - 10=-12$$ To solve this equation, we want to convert it into exponential form. To perform this action, we want to simplify in order to match the format of: $$\log_{a}(x)=k$$ Let's begin by adding 10 to each side of the equation: $$-2\log_{3}(5x + 9)=-2$$ Now, we will divide each side of the equation by -2: $$\log_{3}(5x + 9)=1$$ Now, we can convert it into exponential form: $$3^1=5x + 9$$ Now, we can solve the equation: $$5x + 9=3$$ $$5x=-6$$ $$x=-\frac{6}{5}$$ Example 5: Solve each equation. $$\log_{6}(x + 4) - 4=-2$$ Let's add 4 to each side to start: $$\log_{6}(x + 4)=2$$ Now, let's convert it into exponential form: $$6^{2}=x + 4$$ $$x + 4=36$$ Now, we can solve the equation: $$x=32$$

Example 4: Solve each equation. $$-2\log_{3}(5x + 9) - 10=-12$$ To solve this equation, we want to convert it into exponential form. To perform this action, we want to simplify in order to match the format of: $$\log_{a}(x)=k$$ Let's begin by adding 10 to each side of the equation: $$-2\log_{3}(5x + 9)=-2$$ Now, we will divide each side of the equation by -2: $$\log_{3}(5x + 9)=1$$ Now, we can convert it into exponential form: $$3^1=5x + 9$$ Now, we can solve the equation: $$5x + 9=3$$ $$5x=-6$$ $$x=-\frac{6}{5}$$ Example 5: Solve each equation. $$\log_{6}(x + 4) - 4=-2$$ Let's add 4 to each side to start: $$\log_{6}(x + 4)=2$$ Now, let's convert it into exponential form: $$6^{2}=x + 4$$ $$x + 4=36$$ Now, we can solve the equation: $$x=32$$

## Properties of Logarithms

As we go deeper into the topic of logarithms, we will encounter the various properties of logarithms. Here, we will just focus on two specific properties that are very useful. For any a > 0, a ≠ 1

$$\log_{a}(a)=1$$ $$\log_{a}(1)=0$$ In exponential form: $$a^1=a$$ $$a^0=1$$ Let's look at an example.

Example 6: Evaluate each. $$\log_{14}(14)$$ Using our above rule, we can say this logarithm is equal to 1. We know that 14 raised to the power of 1 is 14. $$\log_{14}(14)=1$$ Example 7: Evaluate each. $$\log_{103}(1)$$ Since 1 is the argument (part inside of the parentheses) of our logarithm, and our base is a positive real number (that is not 1), we can say this logarithm is equal to 0. We know that 103 raised to the power of 0 is 1. $$\log_{103}(1)=0$$

$$\log_{a}(a)=1$$ $$\log_{a}(1)=0$$ In exponential form: $$a^1=a$$ $$a^0=1$$ Let's look at an example.

Example 6: Evaluate each. $$\log_{14}(14)$$ Using our above rule, we can say this logarithm is equal to 1. We know that 14 raised to the power of 1 is 14. $$\log_{14}(14)=1$$ Example 7: Evaluate each. $$\log_{103}(1)$$ Since 1 is the argument (part inside of the parentheses) of our logarithm, and our base is a positive real number (that is not 1), we can say this logarithm is equal to 0. We know that 103 raised to the power of 0 is 1. $$\log_{103}(1)=0$$

## Graphing a Logarithmic Function

$$f(x)=\log_a(x), a > 1$$

f(x) = log

Example 8: Sketch the graph of each.

$$g(x)=\log_{3}(x)$$ To graph this, we can place it in exponential form. We just need to plot enough ordered pairs to get a good idea of the shape of the graph. $$x=3^y$$

We could have also graphed our last problem using a different approach. In our lesson on graphing exponential functions, we sketched the graph of f(x) = 3

Example 9: Sketch the graph of each. $$h(x)=\log_3(x - 1) + 1$$ When we think about this graph, based on the previous example g(x) = log

- The graph is continuous and increasing over its entire domain
- The y-axis or x = 0 is a vertical asymptote as x → 0 from the right

- The graph is continuous and decreasing over its entire domain
- the y-axis or x = 0 is a vertical asymptote as x → 0 from the right

f(x) = log

_{a}(x), a > 0, x > 0, a ≠ 1- (1,0) is on the graph
- if a > 1, the graph rises from left to right
- if 0 < a < 1, the graph falls from left to right
- The graph approaches the y-axis but does not touch it. It forms an asymptote.
- The domain consists of all positive real numbers or the interval: (0, ∞)
- The range consists of all real numbers or the interval: (-∞, ∞)
- The points (1/a, -1), (1, 0), and (a, 1) are on the graph

Example 8: Sketch the graph of each.

$$g(x)=\log_{3}(x)$$ To graph this, we can place it in exponential form. We just need to plot enough ordered pairs to get a good idea of the shape of the graph. $$x=3^y$$

x | y | (x, y) |
---|---|---|

1/9 | -2 | (1/9, -2) |

1/3 | -1 | (1/3, -1) |

1 | 0 | (1, 0) |

3 | 1 | (3, 1) |

9 | 2 | (9, 2) |

^{x}. Notice that our two functions are inverses. If we start with f(x) = 3^{x}: $$f(x)=3^x$$ Replace f(x) with y: $$y=3^x$$ Interchange x and y: $$x=3^y$$ Solve for y: $$y=\log_3(x)$$ Replace with inverse notation: $$f^{-1}(x)=\log_3(x)$$ We will also encounter graphs of logarithmic functions that involve various transformations that we have previously studied. Let's look at an example.Example 9: Sketch the graph of each. $$h(x)=\log_3(x - 1) + 1$$ When we think about this graph, based on the previous example g(x) = log

_{3}(x), it will be shifted 1 unit right and 1 unit up.#### Skills Check:

Example #1

Evaluate each. $$\log_{18}\left(\frac{1}{324}\right)$$

Please choose the best answer.

A

$$2$$

B

$$-2$$

C

$$\frac{1}{2}$$

D

$$-4$$

E

$$\frac{1}{4}$$

Example #2

Solve each equation. $$\log_{3}\left(\frac{1}{27}\right)=x$$

Please choose the best answer.

A

$$x=-3$$

B

$$x=2$$

C

$$x=-1$$

D

$$x=5$$

E

$$x=-9$$

Example #3

Solve each equation. $$\log_{14}(196)=x$$

Please choose the best answer.

A

$$x=4$$

B

$$x=-4$$

C

$$x=6$$

D

$$x=2$$

E

$$x=-2$$

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