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 logarithmic equations
- Learn how to graph a logarithmic function

## Converting Between Logarithmic and Exponential Form

In the last lesson, we 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 form. 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

Now, 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. $$-2log_{3}(5x + 9) - 10=-12$$ 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: $$-2log_{3}(5x + 9)=-2$$ 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. $$-2log_{3}(5x + 9) - 10=-12$$ 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: $$-2log_{3}(5x + 9)=-2$$ 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

For any b > 0, b ≠ 1

$$log_{b}(b)=1$$ $$log_{b}(1)=0$$ In exponential form: $$b^1=b$$ $$b^0=1$$ 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. $$log_{14}(14)=1$$

$$log_{b}(b)=1$$ $$log_{b}(1)=0$$ In exponential form: $$b^1=b$$ $$b^0=1$$ 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. $$log_{14}(14)=1$$

## Graphing a Logarithmic Function

When we think about the graph of the logarithmic function:

f(x) = log

Example 7: Sketch the graph of each.

$$y=log_{3}(x)$$ 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. Since we graphed y = 3

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 has a vertical asymptote at x = 0

- The domain consists of all positive real numbers or the interval: (0, ∞)
- The range consists of all real numbers or the interval: (-∞, ∞)

Example 7: Sketch the graph of each.

$$y=log_{3}(x)$$ 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. Since we graphed y = 3

^{x}in the last lesson, we can interchange the ordered pairs. $$x=3^y$$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) |

#### Skills Check:

Example #1

Rewrite each equation in logarithmic form. $$14^{-18}=x$$

Please choose the best answer.

A

$$log_{-18}(x)=14$$

B

$$log_{x}(-18)=14$$

C

$$log_{14}(-18)=x$$

D

$$log_{14}(x)=-18$$

E

$$log_{18}(-x)=-18$$

Example #2

Solve each equation. $$-3\hspace{.1em}log_{8}(-7x + 2) + 10=4$$

Please choose the best answer.

A

$$x=-\frac{62}{7}$$

B

$$x=-56$$

C

$$x=-\frac{13}{5}$$

D

$$x=-7$$

E

$$x=-\frac{33}{56}$$

Example #3

Solve each equation. $$4\hspace{.1em}log_{2}(5x - 1) + 1=13$$

Please choose the best answer.

A

$$x=84$$

B

$$x=\frac{9}{5}$$

C

$$x=\frac{31}{9}$$

D

$$x=\frac{7}{18}$$

E

$$x=-\frac{1}{31}$$

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