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3.3: Logarithmic Functions

Created by: CK-12

Learning objectives

  • Translate numerical and algebraic expressions between exponential and logarithmic form.
  • Evaluate logarithmic functions.
  • Determine the domain of logarithmic functions.
  • Graph logarithmic functions.
  • Solve logarithmic equations.

Introduction

In the previous lesson we examined exponential expressions and functions. Now we will consider another representation for the same relationships involved in exponential expressions and functions.

Consider the function y= 2x. Every x-value of this function is an exponent. Every y-value is a power of 2. As you learned in lesson 1, functions that are one-to-one have inverses that are functions. This is the case with exponential functions. If we take the inverse of y = 2x (by interchanging the domain and range) we obtain this equation: x = 2y . In order to write this equation such that y is expressed as a function of x, we need a different notation.

The solution to this problem is found in the logarithm. John Napier originally introduced the logarithm to 17th century mathematicians as a technique for simplifying complicated calculations. While today’s technology allows us to do most any calculations we could imagine, logarithmic functions continue to be a focus of study in mathematics, as a useful way to work with exponential expressions and functions.

Changing Between Exponential and Logarithmic Expressions

Every exponential expression can be written in logarithmic form. For example, the equation x=2y is written as follows: y=log2x. In general, the equation logbn = a is equivalent to the equation ba=n. That is, b is the base, a is the exponent, and n is the power, or the result you obtain by raising b to the power of a. Notice that the exponential form of an expression emphasizes the power, while the logarithmic form emphasizes the exponent. More simply put, a logarithm (or “log” for short) is an exponent.

We can write any exponential expression in logarithmic form.

Example 1: Rewrite each exponential expression as a log expression.

a.34=81 b.b4x=52

Solution:

a. In order to rewrite an expression, you must identify its base, its exponent, and its power. The 3 is the base, so it is placed as the subscript in the log expression. The 81 is the power, and so it is placed after the “log”. Thus we have: 34=81 is the same as log381=4 .
To read this expression, we say “the logarithm base 3 of 81 equals 4.” This is equivalent to saying “ 3 to the 4th power equals 81.”
b. The b is the base, and the expression 4x is the exponent, so we have:logb52=4x . We say, “log base b of 52, equals 4x.”

We can also express a logarithmic statement in exponential form.

Example 2: Rewrite the logarithmic expressions in exponential form.

a. log10100=2 b. logbw=5

Solution:

a. The base is 10, and the exponent is 2, so we have: 102=100
b. The base is b, and the exponent is 5, so we have: b5=w .

Perhaps the most common example of a logarithm is the Richter scale, which measures the magnitude of an earthquake. The magnitude is actually the logarithm base 10 of the amplitude of the quake. That is, m=log10A . This means that, for example, an earthquake of magnitude 4 is 10 times as strong as an earthquake with magnitude 3. We can see why this is true of we look at the logarithmic and exponential forms of the expressions: An earthquake of magnitude 3 means 3=log10A. The exponential form of this expression is 103=A. Thus the amplitude of the quake is 1,000. Similarly, a quake with magnitude 4 has amplitude 104=10,000. We will return to this example in lesson 3.8.

Evaluating Logarithmic Functions

As noted above, a logarithmic function is the inverse of an exponential function. Consider again the function y=2x and its inverse x=2y. Above, we rewrote the inverse as y=log2x. If we want to emphasize the fact that the log equation represents a function, we can write the equation as f(x)=log2x. To evaluate this function, we choose values of x and then determine the corresponding y values, or function values.

Example 3: Evaluate the function f(x)=log2x for the values:

a.x=2 b.x=1 c.x=-2

Solution:

a. If x=2 , we have:
f(x)=log2x
f(2)=log22
To determine the value of log22, you can ask yourself: “2 to what power equals 2?” Answering this question is often easy if you consider the exponential form: 2?=2
The missing exponent is 1. So we have f(2)-log22=1
b. If x=1 , we have:
f(x)=log2x
f(1)=log21
As we did in (a), we can consider the exponential form: 2?=1. The missing exponent is 0. So we have f(1)=log21=0.
c. If x=-2, we have:
f(x) =log2x
f(-2) = log2-2
Again, consider the exponential form: 2?=-2. There is no such exponent. Therefore f(-2)=log2-2 does not exist.

Example 3c illustrates an important point: there are restrictions on the domain of a logarithmic function. For the function f(x)=log2x, x cannot be a negative number. Therefore we can state the domain of this function as: “the set of all real numbers greater than 0.” Formally, we can write it as a set: \left \{{x \in \mathbb R  |  x>0} \right \}. In general, the domain of a logarithmic function is restricted to those values that will make the argument of the logarithm non-negative.

For example, consider the function f(x)=log3(x-4). If you attempt to evaluate the function for x values of 4 or less, you will find that the function values do not exist. Therefore the domain of the function is \left \{{x \in \mathbb R  |  x>4} \right \}. The domain of a logarithmic function is one of several key issues to consider when graphing.

Graphing Logarithmic Functions

Because the function f(x)=log2x is the inverse of the function g(x)=2x, the graphs of these functions are reflections over the line y = x. The figure below shows the graphs of these two functions:

We can see that the functions are inverses by looking at the graph. For example, the graph of g(x)=2x contains the point (1, 2), while the graph of f(x)=log2x contains the point (2, 1).

Also, note that while that the graph of g(x)=2x is asymptotic to the x-axis, the graph of f(x)=log2x is asymptotic to the y-axis. This behavior of the graphs gives us a visual interpretation of the restricted range of g and the restricted domain of f.

When graphing log functions, it is important to consider x- values across the domain of the function. In particular, we should look at the behavior of the graph as it gets closer and closer to the asymptote. Consider f(x)=log2x for values of x between 0 and 1.

If x=1/2, then f(1/2)=log2(1/2)=-1 because 2-1=1/2
If x=1/4, then f(1/4)=log_2(1/4)=-2 because 2-2=1/4
If x=1/8}, then f(1/8)=log2(1/8)=-3 because 2-3=1/8

From these values you can see that if we choose x values that are closer and closer to 0, the y values decrease (heading towards - \infty !). In terms of the graph, these values show us that the graph gets closer and closer to the y-axis. Formally we say that the vertical asymptote of the graph is x = 0.

Example 4: Graph the function f(x)=log4x and state the domain and range of the function.

Solution: The function f(x)=log4x is the inverse of the function g(x) = 4x. We can sketch a graph of f(x) by evaluating the function for several values of x, or by reflecting the graph of g over the line y = x.

If we choose to plot points, it is helpful to organize the points in a table:
x y=log4x
1/4
1 0
4 1
16 2

The graph is asymptotic to the y-axis, so the domain of f is the set of all real numbers that are greater than 0. We can write this as a set:\left \{{x \in \mathbb R  |  x>0} \right \} . While the graph might look as if it has a horizontal asymptote, it does in fact continue to rise. The range is \mathbb R.

A note about graphing calculators: You can use a graphing calculator to graph logarithmic functions, but many calculators will only allow you to use base 10 or base e. However, after the next lesson you will be able to rewrite any log as a log with base 10 or base e.

In this section we have looked at graphs of logarithmic functions of the form f(x)=logbx. Now we will consider the graphs of other forms of logarithmic equations.

Graphing Logarithmic Functions Using Transformations

As you saw in the previous lesson, you can graph exponential functions by considering the relationships between equations. For example, you can use the graph of f(x)=2x to sketch a graph of g(x)=2x + 3. Every y value of g(x) is the same as a y value of f(x), plus 3. Therefore we can shift the graph of f(x) up 3 units to obtain a graph of g(x).

We can use the same relationships to efficiently graph log functions. Consider again the log function f(x)=log2x. The table below summarizes how we can use the graph of this function to graph other related function.

Equation Relationship to f(x)=log2x Domain
g(x)=log2(x - a), for a > 0 Obtain a graph of g by shifting the graph of f a units to the right. x > a
g(x) = log2(x+a) for a > 0 Obtain a graph of g by shifting the graph of f a units to the left. x > -a
g(x)=log2(x) + a for a > 0 Obtain a graph of g by shifting the graph of f up a units. x > 0
g(x)=log2(x)-a for a > 0 Obtain a graph of g by shifting the graph of f down a units. x > 0
g(x)=alog2(x) for a > 0 Obtain a graph of g by vertically stretching the graph of f by a factor of a. x > 0
g(x)=-alog2(x) , for a > 0 Obtain a graph of g by vertically stretching the graph of f by a factor of a, and by reflecting the graph over the x-axis. x > 0
g(x)=log2(-x) Obtain a graph of g by reflecting the graph of f over the y-axis. x < 0

Example 5: Graph the functions f(x)=log2(x),g(x) = log2(x) + 3, and h(x) = log2(x + 3)

Solution: The graph below shows these three functions together:

Notice that the location of the 3 in the equation makes a difference! When the 3 is added to log2x , the shift is vertical. When the 3 is added to the x, the shift is horizontal. It is also important to remember that adding 3 to the x is a horizontal shift to the left. This makes sense if you consider the function value when x = -3:

h(-3)=log2(-3 + 3)=log20 = undefined

This is the vertical asymptote! To graph these functions, we evaluated them for certain values of x. But what if we want to know what the x value is for a particular y value? This means that we need to solve a logarithmic equation.

Solving Logarithmic Equations

In general, to solve an equation means to find the value(s) of the variable that makes the equation a true statement. To solve log equations, we have to think about what “log” means.

Consider the equation log2x=5 . What is the exponential form of this equation?
The equation log2x=5 means that 25 = x . So the solution to the equation is x = 25 = 32.

We can use this strategy to solve many logarithmic equations.

Example 6: Solve each equation for x:

a.log4x = 3 b. log5(x + 1) = 2 c. 1 + 2log3(x - 5) = 7

Solution: a. Writing the equation in exponential form gives us the solution: x = 43 = 64.

b. Writing the equation in exponential form gives us a new equation: 52 = x + 1.
We can solve this equation for x:
52 = x + 1
25 = x + 1
x = 24
c. First we have to isolate the log expression:
1 + 2log3(x-5) = 7
2log3(x - 5) = 6
log3(x-5) = 3
Now we can solve the equation by rewriting it in exponential form:
log3(x-5) = 3
33 = x - 5
27 = x - 5
x = 32

We can also solve equations in which both sides of the equation contain logs. For example, consider the equation log2(3x-1)=log2(5x - 7). Because the logarithms have the same base (2), the arguments of the log (the expressions 3x - 1 and 5x - 7) must be equal. So we can solve as follows:

log3(3x-1) = log2(5x - 7)
3x - 1 = 5x - 7
+7 |+7
3x + 6 = 5x
-3x
6 = 2x
x = 3

Example 7: Solve for x: log2(9x)=log2(3x + 8)

Solution: The log equation implies that the expressions 9x and 3x + 8 are equal:

log_2(9x) = log_2(3x + 8)
9x = 3x + 8
-3x -3x
6x = 8
x = \frac{8} {6}
x = \frac{4} {3}

Lesson Summary

In this lesson we have defined the logarithmic function as the inverse of the exponential function. When working with logarithms, it helps to keep in mind that a logarithm is an exponent. For example, 3 = log28 and 23 = 8 are two forms of the same numerical relationship among the three numbers 2, 3, and 8. The 2 is the base, the 3 is the exponent, and 8 is the 3rd power of 2.

Because logarithmic functions are the inverses of exponential functions, we can use our knowledge of exponential functions to graph logarithmic functions. You can graph a log function either by reflecting an exponential function over the line y = x, or by evaluating the function and plotting points. In this lesson you learned how to graph “parent” graphs such as y =log2x and y = log4x , as well as how to use these parent graphs to graph more complicated log functions. When graphing, it is important to keep in mind that logarithmic functions have restricted domains. Each graph will have a vertical asymptote.

We can also use our knowledge of exponential relationships to solve logarithmic equations. In this lesson we solved 2 kinds of logarithmic equations. First, we solved equations by rewriting the equations in exponential form. Second, we solved equations in which both sides of the equation contained a log. To solve these equations, we used the following rule:

logb f(x)=logb g(x) → f(x) = g(x) .

Points to Consider

  1. What methods can you use to graph logarithmic functions?
  2. What methods can you use to solve logarithmic equations?
  3. What forms of log equations can you solve using the methods in this lesson? Can you write an equation that cannot be solved using these methods?

Review Questions

Write the exponential statement in logarithmic form.

  1. 32 = 9
  2. z4 = 10
  3. Write the logarithmic statement in exponential form.
  4. log5 25 = 2
  5. log_4{\frac {1}{6}}=-1
  6. Complete the table of values for the function f(x) = log3 x
  7. x y = f(x)
    1/9
    1/3
    1
    3
    9
  8. Use the table above to graph f(x)=log3x. State the domain and range of the function.
  9. Consider g(x) = -log3(x - 2) a. How is the graph of g(x) related to the graph of f(x) = log3x? b. Graph g(x) by transforming the graph of f(x).
  10. Solve each logarithmic equation: a. log3 9x = 4 b. 7 + log2 x = 11 (Hint: subtract 7 from both sides first.)
  11. Solve each logarithmic equation:
  12. a. log5 6x = -1 b. log5 6x = log5(2x + 16) c. log5 6x = log5(3x - 10)
  13. Explain why the equation in 9c has no solution.

Review Answer

  1. 32 = 9
  2. z4 =10
  3. 52 = 25
  4. 6^{-1}= \frac {1}{6}
  5. x y = f(x)
    1/9
    1/3
    1 0
    3 1
    9 2
  6. D: All real numbers > 0 R: All real numbers.
  7. a. The graph of g(x) can be obtained by shifting the graph of f(x) 2 units to the right, and reflecting it over the x- axis. b.
  8. The solutions are:
  9. a. x = 9 b. x = 16
  10. The solutions are:
  11. a. x = 1/30 b. x = 4 c. no solution
  12. When we solve 6x=3x-10 we find that x=-10/3, a value outside of the domain. Because there is no other x value that satisfies the equation, there is no solution.

Vocabulary

Argument
The expression “inside” a logarithmic expression. The argument represents the “power” in the exponential relationship.
Asymptote
An asymptote is line whose distance to a given curve tends to zero. An asymptote may or may not intersect its associated curve.
Domain
The domain of a function is the set of all values of the independent variable (x) for which the function is defined.
Evaluate
To evaluate a function is to identify a function value (y) for a given value of the independent variable (x).
Function
A function is a relation between a domain (set of x values) and range (set of y values) in which every element of the domain is paired with one and only one element of the range. A function that is “one to one” is a function in which every element of the domain is paired with exactly one y value.
Logarithm
The exponent of the power to which a base number must be raised to equal a given number.
Range
The range of a function is the set of all function values, or values of the dependent variable (y).

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