Graphs of Logarithmic Functions

Learning Objectives

Find the domain and range of a relation and a function. (IA 3.5.1)
Graph Logarithmic functions. (IA 10.3.3)

Objective 1: Find the domain and range of a relation and a function. (IA 3.5.1)

Example 1

Find the domain and range of a relation and a function.

ⓐ
ⓑ
ⓒ
Find the domain of the function $f (x) = \frac{5}{x - 2}$
ⓓ
Find the domain of the function $f (x) = \log_{2} (x - 5)$ .

Solution

ⓐ
The set of points on the graph is ${(- 4, - 2), (- 2, - 1), (- 1, 1), (1, 2)}$
The Domain is the set of all x-coordinates: ${- 4, - 2, - 1, 1}$
The Range is the set of all y-coordinates: ${- 2, - 1, 1}$
Notice that even though y-coodinate of 1 appears twice, we only list it once.
ⓑ
Domain: $(- \infty, \infty)$
Range: $[-2, \infty)$
Notice that $-2$ is included because the point $(3, -2)$ is on the graph of a function.
ⓒ
A function is not defined when the denominator is zero. We need to set the denominator equal zero and exclude this value(s) from the domain.
$x - 2 = 0, x = 2,$ Domain $(- \infty, 2) \cup (2, \infty)$
Notice that 2 is excluded from the domain because the function is not defined at $x = 2$
ⓓ
From the definition of the logarithmic function $f (x) = \log_{a} x$ we know that $x > 0$
To find domain of $f (x) = \log_{2} (x - 5$ , we need to set up and solve inequality.
$x - 5 > 0$ ,
$x > 5)$ Domain: $(5, \infty)$

Practice Makes Perfect

Find the domain and range of a relation and a function.

Find the domain and range of a relation. A scatter plot displaying four blue points on a Cartesian coordinate plane. The points are located at (-3, 1), (-1, -2), (2, 2), and (2, 4).

Find the domain and the range of the function graphed. Use interval notation. A graph illustrates an increasing logarithmic function on a Cartesian plane, spanning x-values from -5 to 9 and y-values from -6 to 6. The curve passes through points like (-3, 0), (-2, 1), (0, 2), and (4, 3).

Find the domain of the function $f (x) = \log_{2} (x + 4)$ . Notice: this is the same function that was graphed in question 2.

Objective 2: Graph Logarithmic functions. (IA 10.3.3)

To graph a logarithmic function $y = \log_{a} x$ , it is easiest to convert the equation to its exponential form, $x = a^{y}$ . Generally, when we look for ordered pairs for the graph of a function, we usually choose an x-value and then determine its corresponding y-value. In this case you may find it easier to choose y-values and then determine its corresponding x-value.

Example 2

Graph Logarithmic functions.

Graph $y = {log}_{2} x .$

Solution

To graph the function, we will first rewrite the logarithmic equation, $y = {log}_{2} x,$ in exponential form, $2^{y} = x .$

We will use point plotting to graph the function. It will be easier to start with values of y and then get x.

This table has three columns and seven rows. The first row is a header row and it reads y, 2 to the y power equals x, and (x, y). In the first column below y we have negative 2, negative 1, 0, 1, 2, and 3. In the second column below 2 to the y power equals x we have 2 to the negative 2 power equals 1 over 2 squared which equals 1 over 4, 2 to the negative 1 power equals 1 over 2 to the first power which equals 1 over 2, 2 to the negative 0 power equals 2, 2 to the 1 power equals 2, 2 squared equals 4, and 2 cubed equals 8. In the third column below (x, y) we have (1 over 4, 2), (1 over 2, negative 1), (1, 0), (2, 1), (4, 2), and (8, 3).
$y$	$2^{y} = x$	$(x, y)$
$−2$	$2^{−2} = \frac{1}{2^{2}} = \frac{1}{4}$	$(\frac{1}{4}, 2)$
$−1$	$2^{−1} = \frac{1}{2^{1}} = \frac{1}{2}$	$(\frac{1}{2}, −1)$
0	$2^{0} = 1$	$(1, 0)$
1	$2^{1} = 2$	$(2, 1)$
2	$2^{2} = 4$	$(4, 2)$
3	$2^{3} = 8$	$(8, 3)$

This figure shows the logarithmic curve going through the points (1 over 2, negative 1), (1, 0), and (2, 1).

Practice Makes Perfect

Graph Logarithmic functions

Graph

y = {log}_{3} x

and

y = {log}_{5} x

in the same coordinate system.

.
$y$	$3^{y} = x$	$(x, y)$






$y$	$5^{y} = x$	$(x, y)$

A blank Cartesian coordinate system with labeled x and y axes ranging from -10 to 10, complete with grid lines.

Graph

y = \log_{1 / 3} x

.
$y$	${(\frac{1}{3})}^{y} = x$	$(x, y)$

This figure shows the logarithmic curve going through the points (1 over a, negative 1), (1, 0), and (a, 1).
Do the graphs of $y = \log_{2} x$ , $y = \log_{3} x$ , and $y = \log_{5} x$ have the shape we expect from a logarithmic function where $a > 0$ ? (Remember a is the base of the log function)

Is there a point they all share? Why does this make sense?

Do they all have a point $(a, 1)$ ? Why does this make sense?

Do they all have a point $(\frac{1}{a}, -1)$ ? Why does this make sense?

Do they all have the same vertical asymptote? What is the equation of the vertical asymptote?

Do they all have the same domain? Write the domain in the interval notation.

Do they all have the same range? Write the range in the interval notation.

Finding the Domain of a Logarithmic Function

Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined.

Recall that the exponential function is defined as $y = b^{x}$ for any real number $x$ and constant $b > 0,$ $b \neq 1,$ where

The domain of $y$ is $(- \infty, \infty) .$
The range of $y$ is $(0, \infty) .$

In the last section we learned that the logarithmic function $y = \log_{b} (x)$ is the inverse of the exponential function $y = b^{x} .$ So, as inverse functions:

The domain of $y = \log_{b} (x)$ is the range of $y = b^{x} :$ $(0, \infty) .$
The range of $y = \log_{b} (x)$ is the domain of $y = b^{x} :$ $(- \infty, \infty) .$

Transformations of the parent function $y = \log_{b} (x)$ behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, stretches, compressions, and reflections.

In Graphs of Exponential Functions we saw that certain transformations can change the range of $y = b^{x} .$ Similarly, applying transformations to the parent function $y = \log_{b} (x)$ can change the domain. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists only of positive real numbers. That is, the argument of the logarithmic function must be greater than zero.

For example, consider $f (x) = \log_{4} (2 x - 3) .$ This function is defined for any values of $x$ such that the argument, in this case $2 x - 3,$ is greater than zero. To find the domain, we set up an inequality and solve for $x :$

\begin{array}{l} 2 x - 3 > 0 & Show the argument greater than zero . \\ 2 x > 3 & Add 3 . \\ x > 1.5 \begin{matrix}  \end{matrix} & Divide by 2 . \end{array}

In interval notation, the domain of $f (x) = \log_{4} (2 x - 3)$ is $(1.5, \infty) .$

Example 3

Identifying the Domain of a Logarithmic Shift

What is the domain of $f (x) = \log_{2} (x + 3) ?$

Solution

The logarithmic function is defined only when the input is positive, so this function is defined when $x + 3 > 0.$ Solving this inequality,

\begin{array}{l} x + 3 > 0 & The input must be positive . \\ x > - 3 \begin{matrix}  \end{matrix} & Subtract 3 . \end{array}

The domain of $f (x) = \log_{2} (x + 3)$ is $(- 3, \infty) .$

Example 4

Identifying the Domain of a Logarithmic Shift and Reflection

What is the domain of $f (x) = \log (5 - 2 x) ?$

Solution

The logarithmic function is defined only when the input is positive, so this function is defined when $5 - 2 x > 0 .$ Solving this inequality,

\begin{array}{l} 5 - 2 x > 0 & The input must be positive . \\ - 2 x > - 5 & Subtract 5. \\ x < \frac{5}{2} \begin{matrix}  \end{matrix} & Divide by - 2 and switch the inequality . \end{array}

The domain of $f (x) = \log (5 - 2 x)$ is $(- \infty, \frac{5}{2}) .$

Graphing Logarithmic Functions

Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function $y = \log_{b} (x)$ along with all its transformations: shifts, stretches, compressions, and reflections.

We begin with the parent function $y = \log_{b} (x) .$ Because every logarithmic function of this form is the inverse of an exponential function with the form $y = b^{x},$ their graphs will be reflections of each other across the line $y = x .$ To illustrate this, we can observe the relationship between the input and output values of $y = 2^{x}$ and its equivalent $x = \log_{2} (y)$ in Table 1.

Table 1 Three rows and eight columns. The first row is labeled, “x”, the second row is labeled, “y=2^x”, and the third row is labeled, “log_2(y)=x”. Reading the columns as ordered pairs, we have the following values for the second row: (-3, 1/8), (-2, 1/4), (-1, 1/2), (0, 1), (1, 2), (2, 4), and (3, 8). For the third row: (1/8, -3), (1/4, -2), (1/2, -1), (1, 0), (2, 1), (4, 2), and (8, 3).
$x$	$- 3$	$- 2$	$- 1$	$0$	$1$	$2$	$3$
$2^{x} = y$	$\frac{1}{8}$	$\frac{1}{4}$	$\frac{1}{2}$	$1$	$2$	$4$	$8$
$\log_{2} (y) = x$	$- 3$	$- 2$	$- 1$	$0$	$1$	$2$	$3$

Using the inputs and outputs from Table 1, we can build another table to observe the relationship between points on the graphs of the inverse functions $f (x) = 2^{x}$ and $g (x) = \log_{2} (x) .$ See Table 2.

Table 2 Two rows and eight columns. The first row is labeled, “f(x)=2^x”, with the following values: (-3, 1/8), (-2, 1/4), (-1, 1/2), (0, 1), (1, 2), (2, 4), and (3, 8). The second row is labeled, “g(x)=log_2(x)”, with the following values: (1/8, -3), (1/4, -2), (1/2, -1), (1, 0), (2, 1), (4, 2), and (8, 3).
$f (x) = 2^{x}$	$(- 3, \frac{1}{8})$	$(- 2, \frac{1}{4})$	$(- 1, \frac{1}{2})$	$(0, 1)$	$(1, 2)$	$(2, 4)$	$(3, 8)$
$g (x) = \log_{2} (x)$	$(\frac{1}{8}, - 3)$	$(\frac{1}{4}, - 2)$	$(\frac{1}{2}, - 1)$	$(1, 0)$	$(2, 1)$	$(4, 2)$	$(8, 3)$

As we’d expect, the x- and y-coordinates are reversed for the inverse functions. Figure 2 shows the graph of $f$ and $g .$

Graph of two functions, f(x)=2^x and g(x)=log_2(x), with the line y=x denoting the axis of symmetry. — Figure 2 Notice that the graphs of $f (x) = 2^{x}$ and $g (x) = \log_{2} (x)$ are reflections about the line $y = x .$

Observe the following from the graph:

$f (x) = 2^{x}$ has a y-intercept at $(0, 1)$ and $g (x) = \log_{2} (x)$ has an x- intercept at $(1, 0) .$
The domain of $f (x) = 2^{x},$ $(- \infty, \infty),$ is the same as the range of $g (x) = \log_{2} (x) .$
The range of $f (x) = 2^{x},$ $(0, \infty),$ is the same as the domain of $g (x) = \log_{2} (x) .$

Characteristics of the Graph of the Parent Function, $f (x) = \log_{b} (x) :$

For any real number $x$ and constant $b > 0,$ $b \neq 1,$ we can see the following characteristics in the graph of $f (x) = \log_{b} (x) :$

one-to-one function
vertical asymptote: $x = 0$
domain: $(0, \infty)$
range: $(- \infty, \infty)$
x-intercept: $(1, 0)$ and key point $(b, 1)$
y-intercept: none
increasing if $b > 1$
decreasing if $0 < b < 1$

See Figure 3.

Figure 3

Figure 4 shows how changing the base $b$ in $f (x) = \log_{b} (x)$ can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (Note: recall that the function $\ln (x)$ has base $e \approx 2 . 718.)$

Figure 4 The graphs of three logarithmic functions with different bases, all greater than 1.

Example 5

Graphing a Logarithmic Function with the Form f(x) = log_b(x).

Graph $f (x) = \log_{5} (x) .$ State the domain, range, and asymptote.

Solution

Before graphing, identify the behavior and key points for the graph.

Since $b = 5$ is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote $x = 0,$ and the right tail will increase slowly without bound.
The x-intercept is $(1, 0) .$
The key point $(5, 1)$ is on the graph.
We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points (see Figure 5).

Graph of f(x)=log_5(x) with labeled points at (1, 0) and (5, 1). The y-axis is the asymptote. — Figure 5

The domain is $(0, \infty),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 0.$

Graphing Transformations of Logarithmic Functions

As we mentioned in the beginning of the section, transformations of logarithmic graphs behave similarly to those of other parent functions. We can shift, stretch, compress, and reflect the parent function $y = \log_{b} (x)$ without loss of shape.

Graphing a Horizontal Shift of f(x) = log_b(x)

When a constant $c$ is added to the input of the parent function $f (x) = l o g_{b} (x),$ the result is a horizontal shift $c$ units in the opposite direction of the sign on $c .$ To visualize horizontal shifts, we can observe the general graph of the parent function $f (x) = \log_{b} (x)$ and for $c > 0$ alongside the shift left, $g (x) = \log_{b} (x + c),$ and the shift right, $h (x) = \log_{b} (x - c) .$ See Figure 6.

Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=log_b(x+c) is the translation function with an asymptote at x=-c. This shows the translation of shifting left. — Figure 6

Example 6

Graphing a Horizontal Shift of the Parent Function y = log_b(x)

Sketch the horizontal shift $f (x) = \log_{3} (x - 2)$ alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.

Solution

Since the function is $f (x) = \log_{3} (x - 2),$ we notice $x + (- 2) = x - 2.$

Thus $c = - 2,$ so $c < 0.$ This means we will shift the function $f (x) = \log_{3} (x)$ right 2 units.

The vertical asymptote is $x = - (- 2)$ or $x = 2.$

Consider the three key points from the parent function, $(\frac{1}{3}, −1),$ $(1, 0),$ and $(3, 1) .$

The new coordinates are found by adding 2 to the $x$ coordinates.

Label the points $(\frac{7}{3}, −1),$ $(3, 0),$ and $(5, 1) .$

The domain is $(2, \infty),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 2.$

Graph of two functions. The parent function is y=log_3(x), with an asymptote at x=0 and labeled points at (1/3, -1), (1, 0), and (3, 1).The translation function f(x)=log_3(x-2) has an asymptote at x=2 and labeled points at (3, 0) and (5, 1). — Figure 7

Graphing a Vertical Shift of f(x) = log_b(x)

When a constant $d$ is added to the parent function $f (x) = \log_{b} (x),$ the result is a vertical shift $d$ units in the direction of the sign on $d .$ To visualize vertical shifts, we can observe the general graph of the parent function $f (x) = \log_{b} (x)$ alongside the shift up, $g (x) = \log_{b} (x) + d$ and the shift down, $h (x) = \log_{b} (x) - d .$ See Figure 8.

Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=log_b(x)+d is the translation function with an asymptote at x=0. This shows the translation of shifting up. Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=log_b(x)-d is the translation function with an asymptote at x=0. This shows the translation of shifting down. — Figure 8

Example 7

Graphing a Vertical Shift of the Parent Function f(x) = log_b(x)

Sketch a graph of $f (x) = \log_{3} (x) - 2$ alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.

Solution

Since the function is $f (x) = \log_{3} (x) - 2,$ we will notice $d = - 2.$ Thus $d < 0.$

This means we will shift the function $f (x) = \log_{3} (x)$ down 2 units.

The vertical asymptote is $x = 0.$

Consider the three key points from the parent function, $(\frac{1}{3}, −1),$ $(1, 0),$ and $(3, 1) .$

The new coordinates are found by subtracting 2 from the y coordinates.

Label the points $(\frac{1}{3}, −3),$ $(1, −2),$ and $(3, −1) .$

The domain is $(0, \infty),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 0.$

Graphing Stretches and Compressions of f(x) = log_b(x)

When the parent function $f (x) = \log_{b} (x)$ is multiplied by a constant $a > 0,$ the result is a vertical stretch or compression of the original graph. To visualize stretches and compressions, we set $a > 1$ and observe the general graph of the parent function $f (x) = \log_{b} (x)$ alongside the vertical stretch, $g (x) = a \log_{b} (x)$ and the vertical compression, $h (x) = \frac{1}{a} \log_{b} (x) .$ See Figure 10.

Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=alog_b(x) when a>1 is the translation function with an asymptote at x=0. The graph note the intersection of the two lines at (1, 0). This shows the translation of a vertical stretch. — Figure 10

Example 8

Graphing a Stretch or Compression of the Parent Function f(x) = log_b(x)

Sketch a graph of $f (x) = 2 \log_{4} (x)$ alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.

Solution

Since the function is $f (x) = 2 \log_{4} (x),$ we will notice $a = 2.$

This means we will stretch the function $f (x) = \log_{4} (x)$ by a factor of 2.

The vertical asymptote is $x = 0.$

Consider the three key points from the parent function, $(\frac{1}{4}, −1),$ $(1, 0),$ and $(4, 1) .$

The new coordinates are found by multiplying the $y$ coordinates by 2.

Label the points $(\frac{1}{4}, −2),$ $(1, 0),$ and $(4, 2) .$

The domain is $(0, \infty),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 0.$ See Figure 11.

Graph of two functions. The parent function is y=log_4(x), with an asymptote at x=0 and labeled points at (1, 0), and (4, 1).The translation function f(x)=2log_4(x) has an asymptote at x=0 and labeled points at (1, 0) and (2, 1). — Figure 11

The domain is $(0, \infty),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 0.$

Example 9

Combining a Shift and a Stretch

Sketch a graph of $f (x) = 5 \log (x + 2) .$ State the domain, range, and asymptote.

Solution

Remember: what happens inside parentheses happens first. First, we move the graph left 2 units, then stretch the function vertically by a factor of 5, as in Figure 12. The vertical asymptote will be shifted to $x = −2.$ The x-intercept will be $(−1, 0) .$ The domain will be $(−2, \infty) .$ Two points will help give the shape of the graph: $(−1, 0)$ and $(8, 5).$ We chose $x = 8$ as the x-coordinate of one point to graph because when $x = 8,$ $x + 2 = 10,$ the base of the common logarithm.

Graph of three functions. The parent function is y=log(x), with an asymptote at x=0. The first translation function y=5log(x+2) has an asymptote at x=-2. The second translation function y=log(x+2) has an asymptote at x=-2. — Figure 12

The domain is $(- 2, \infty),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = - 2.$

Graphing Reflections of f(x) = log_b(x)

When the parent function $f (x) = \log_{b} (x)$ is multiplied by $−1,$ the result is a reflection about the x-axis. When the input is multiplied by $−1,$ the result is a reflection about the y-axis. To visualize reflections, we restrict $b > 1,$ and observe the general graph of the parent function $f (x) = \log_{b} (x)$ alongside the reflection about the x-axis, $g (x) = {−log}_{b} (x)$ and the reflection about the y-axis, $h (x) = \log_{b} (- x) .$

Graph of two functions. The parent function is f(x)=log_b(x), with an asymptote at x=0 and g(x)=-log_b(x) when b>1 is the translation function with an asymptote at x=0. The graph note the intersection of the two lines at (1, 0). This shows the translation of a reflection about the x-axis. — Figure 13

How To

Given a logarithmic function with the parent function $f (x) = \log_{b} (x),$ graph a translation.

Table 3 The first column gives the following instructions of graphing a translation of f(x)=-log_b(x) with the parent function being f(x)=log_b(x): 1. Draw the vertical asymptote, x=0; 2. Plot the x-intercept, (1, 0); 3. Reflect the graph of the parent function f(x)=log_b(x) about the x-axis; 4. Draw a smooth curve through the points; 5. State the domain, (0, infinity), the range, (-infinity, infinity), and the vertical asymptote x=0. The second column gives the following instructions of graphing a translation of f(x)=log_b(-x) with the parent function being f(x)=log_b(x): 1. Draw the vertical asymptote, x=0; 2. Plot the x-intercept, (-1, 0); 3. Reflect the graph of the parent function f(x)=log_b(x) about the y-axis; 4. Draw a smooth curve through the points; 5. State the domain, (-infinity, 0), the range, (-infinity, infinity), and the vertical asymptote x=0.
$If f (x) = - \log_{b} (x)$	$If f (x) = \log_{b} (- x)$
1. Draw the vertical asymptote, $x = 0.$	1. Draw the vertical asymptote, $x = 0.$
2. Plot the x-intercept, $(1, 0) .$	2. Plot the x-intercept, $(1, 0) .$
3. Reflect the graph of the parent function $f (x) = \log_{b} (x)$ about the x-axis.	3. Reflect the graph of the parent function $f (x) = \log_{b} (x)$ about the y-axis.
4. Draw a smooth curve through the points.	4. Draw a smooth curve through the points.
5. State the domain, (0, ∞), the range, (−∞, ∞), and the vertical asymptote $x = 0$ .	5. State the domain, (−∞, 0) the range, (−∞, ∞) and the vertical asymptote $x = 0.$

Example 10

Graphing a Reflection of a Logarithmic Function

Sketch a graph of $f (x) = \log (- x)$ alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.

Solution

Before graphing $f (x) = \log (- x),$ identify the behavior and key points for the graph.

Since $b = 10$ is greater than one, we know that the parent function is increasing. Since the input value is multiplied by $−1,$ $f$ is a reflection of the parent graph about the y-axis. Thus, $f (x) = \log (- x)$ will be decreasing as $x$ moves from negative infinity to zero, and the right tail of the graph will approach the vertical asymptote $x = 0.$
The x-intercept is $(−1, 0) .$
We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points.

Graph of two functions. The parent function is y=log(x), with an asymptote at x=0 and labeled points at (1, 0), and (10, 1).The translation function f(x)=log(-x) has an asymptote at x=0 and labeled points at (-1, 0) and (-10, 1). — Figure 14

The domain is $(- \infty, 0),$ the range is $(- \infty, \infty),$ and the vertical asymptote is $x = 0.$

Example 11

Approximating the Solution of a Logarithmic Equation

Solve $4 \ln (x) + 1 = - 2 \ln (x - 1)$ graphically. Round to the nearest thousandth.

Solution

Press [Y=] and enter $4 \ln (x) + 1$ next to Y₁=. Then enter $- 2 \ln (x - 1)$ next to Y₂=. For a window, use the values 0 to 5 for $x$ and –10 to 10 for $y .$ Press [GRAPH]. The graphs should intersect somewhere a little to right of $x = 1.$

For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 1.3385297. (Your answer may be different if you use a different window or use a different value for Guess?) So, to the nearest thousandth, $x \approx 1.339.$

Summarizing Translations of the Logarithmic Function

Now that we have worked with each type of translation for the logarithmic function, we can summarize each in Table 4 to arrive at the general equation for translating exponential functions.

Table 4 Titled, Transformations of the Parent Function f(x)=lob_b(x)”. The first column is labeled, “Transformation”, and the second column is labeled, “Form”. The first transformation is a horizontal and vertical shift with c units to the left and d units up with the form f(x)=log_b(x+c)+d. The second transformation is a stretch and compression. It is a stretch is |a|>1 and a compression is 0<|a|<1. Note its form is f(x)=alog_b(x). The third transformation is a reflection about the x-axis with the form f(x) = -log_b(x). The fourth transformation is a reflection about the y-axis with the form f(x)=log_b(-x). The general equation for all transformation is f(x)=alog_b(x+c)+d.
Transformations of the Parent Function $y = \log_{b} (x)$
Transformation	Form
Shift Horizontally $c$ units to the left Vertically $d$ units up	$y = \log_{b} (x + c) + d$
Stretch and Compress Stretch if $\| a \| > 1$ Compression if $\| a \| < 1$	$y = a \log_{b} (x)$
Reflect about the x-axis	$y = - \log_{b} (x)$
Reflect about the y-axis	$y = \log_{b} (- x)$
General equation for all translations	$y = a \log_{b} (x + c) + d$

Example 12

Finding the Vertical Asymptote of a Logarithm Graph

What is the vertical asymptote of $f (x) = −2 \log_{3} (x + 4) + 5 ?$

Solution

The vertical asymptote is at $x = - 4.$

Analysis

The coefficient, the base, and the upward translation do not affect the asymptote. The shift of the curve 4 units to the left shifts the vertical asymptote to $x = −4.$

Example 13

Finding the Equation from a Graph

Find a possible equation for the common logarithmic function graphed in Figure 15.

Graph of a logarithmic function with a vertical asymptote at x=-2, has been vertically reflected, and passes through the points (-1, 1) and (2, -1). — Figure 15

Solution

This graph has a vertical asymptote at $x = –2$ and has been vertically reflected. We do not know yet the vertical shift or the vertical stretch. We know so far that the equation will have form:

f (x) = - a \log (x + 2) + k

It appears the graph passes through the points $(–1, 1)$ and $(2, –1) .$ Substituting $(–1, 1),$

\begin{array}{l} 1 = - a \log (−1 + 2) + k & Substitute (−1, 1) . \\ 1 = - a \log (1) + k & Arithmetic . \\ 1 = k & log(1) = 0. \end{array}

Next, substituting in $(2, –1)$ ,

\begin{array}{l} - 1 = - a \log (2 + 2) + 1 & Plug in (2, −1) . \\ - 2 = - a \log (4) & Arithmetic . \\ a = \frac{2}{\log (4)} & Solve for a . \end{array}

This gives us the equation $f (x) = - \frac{2}{\log (4)} \log (x + 2) + 1.$

Analysis

We can verify this answer by comparing the function values in Table 5 with the points on the graph in Figure 15.

Table 5 ..
$x$	−1	0	1	2	3
$f (x)$	1	0	−0.58496	−1	−1.3219
$x$	4	5	6	7	8
$f (x)$	−1.5850	−1.8074	−2	−2.1699	−2.3219

Key Equations

...
General Form for the Translation of the Parent Logarithmic Function $f (x) = \log_{b} (x)$	$f (x) = a \log_{b} (x + c) + d$

Key Concepts

To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for $x .$ See Example 3 and Example 4
The graph of the parent function f(x)= log b ( x ) has an x-intercept at ( 1,0 ), domain ( 0,∞ ), range ( −∞,∞ ), vertical asymptote x=0, and
- if $b > 1,$ the function is increasing.
- if $0 < b < 1,$ the function is decreasing.
See Example 5.
The equation f(x)= log b ( x+c ) shifts the parent function y= log b ( x ) horizontally
- left $c$ units if $c > 0.$
- right $c$ units if $c < 0.$
See Example 6.
The equation f(x)= log b ( x )+d shifts the parent function y= log b ( x ) vertically
- up $d$ units if $d > 0.$
- down $d$ units if $d < 0.$
See Example 7.
For any constant a>0, the equation f(x)=a log b ( x )
- stretches the parent function $y = \log_{b} (x)$ vertically by a factor of $a$ if $| a | > 1.$
- compresses the parent function $y = \log_{b} (x)$ vertically by a factor of $a$ if $| a | < 1.$
See Example 8 and Example 9.
When the parent function y= log b ( x ) is multiplied by −1, the result is a reflection about the x-axis. When the input is multiplied by −1, the result is a reflection about the y-axis.
- The equation $f (x) = - \log_{b} (x)$ represents a reflection of the parent function about the x-axis.
- The equation $f (x) = \log_{b} (- x)$ represents a reflection of the parent function about the y-axis.
See Example 10.
- A graphing calculator may be used to approximate solutions to some logarithmic equations See Example 11.
All translations of the logarithmic function can be summarized by the general equation $f (x) = a \log_{b} (x + c) + d .$ See Table 4.
Given an equation with the general form $f (x) = a \log_{b} (x + c) + d,$ we can identify the vertical asymptote $x = - c$ for the transformation. See Example 12.
Using the general equation $f (x) = a \log_{b} (x + c) + d,$ we can write the equation of a logarithmic function given its graph. See Example 13.

Section Exercises

Verbal

Exercise 1

The inverse of every logarithmic function is an exponential function and vice-versa. What does this tell us about the relationship between the coordinates of the points on the graphs of each?

Solution

Since the functions are inverses, their graphs are mirror images about the line $y = x .$ So for every point $(a, b)$ on the graph of a logarithmic function, there is a corresponding point $(b, a)$ on the graph of its inverse exponential function.

Exercise 2

What type(s) of translation(s), if any, affect the range of a logarithmic function?

Exercise 3

What type(s) of translation(s), if any, affect the domain of a logarithmic function?

Solution

Shifting the function right or left and reflecting the function about the y-axis will affect its domain.

Exercise 4

Consider the general logarithmic function $f (x) = \log_{b} (x) .$ Why can’t $x$ be zero?

Exercise 5

Does the graph of a general logarithmic function have a horizontal asymptote? Explain.

Solution

No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.

Algebraic

For the following exercises, state the domain and range of the function.

Exercise 6

$f (x) = \log_{3} (x + 4)$

Exercise 7

$h (x) = \ln (\frac{1}{2} - x)$

Solution

Domain: $(- \infty, \frac{1}{2});$ Range: $(- \infty, \infty)$

Exercise 8

$g (x) = \log_{5} (2 x + 9) - 2$

Exercise 9

$h (x) = \ln (4 x + 17) - 5$

Solution

Domain: $(- \frac{17}{4}, \infty);$ Range: $(- \infty, \infty)$

Exercise 10

$f (x) = \log_{2} (12 - 3 x) - 3$

For the following exercises, state the domain and the vertical asymptote of the function.

Exercise 11

$f (x) = \log_{b} (x - 5)$

Solution

Domain: $(5, \infty);$ Vertical asymptote: $x = 5$

Exercise 12

$g (x) = \ln (3 - x)$

Exercise 13

$f (x) = \log (3 x + 1)$

Solution

Domain: $(- \frac{1}{3}, \infty);$ Vertical asymptote: $x = - \frac{1}{3}$

Exercise 14

$f (x) = 3 \log (- x) + 2$

Exercise 15

$g (x) = - \ln (3 x + 9) - 7$

Solution

Domain: $(- 3, \infty);$ Vertical asymptote: $x = - 3$

For the following exercises, state the domain, vertical asymptote, and end behavior of the function.

Exercise 16

$f (x) = \ln (2 - x)$

Exercise 17

$f (x) = \log (x - \frac{3}{7})$

Solution

Domain: $(\frac{3}{7}, \infty)$ ;
Vertical asymptote: $x = \frac{3}{7}$ ; End behavior: as $x \to {(\frac{3}{7})}^{+}, f (x) \to - \infty$ and as $x \to \infty, f (x) \to \infty$

Exercise 18

$h (x) = - \log (3 x - 4) + 3$

Exercise 19

$g (x) = \ln (2 x + 6) - 5$

Solution

Domain: $(- 3, \infty)$ ; Vertical asymptote: $x = - 3$ ;
End behavior: as $x \to - 3^{+}$ , $f (x) \to - \infty$ and as $x \to \infty$ , $f (x) \to \infty$

Exercise 20

$f (x) = \log_{3} (15 - 5 x) + 6$

For the following exercises, state the domain, range, and x- and y-intercepts, if they exist. If they do not exist, write DNE.

Exercise 21

$h (x) = \log_{4} (x - 1) + 1$

Solution

Domain: $(1, \infty);$ Range: $(- \infty, \infty);$ Vertical asymptote: $x = 1;$ x-intercept: $(\frac{5}{4}, 0);$ y-intercept: DNE

Exercise 22

$f (x) = \log (5 x + 10) + 3$

Exercise 23

$g (x) = \ln (- x) - 2$

Solution

Domain: $(- \infty, 0);$ Range: $(- \infty, \infty);$ Vertical asymptote: $x = 0;$ x-intercept: $(- e^{2}, 0);$ y-intercept: DNE

Exercise 24

$f (x) = \log_{2} (x + 2) - 5$

Exercise 25

$h (x) = 3 \ln (x) - 9$

Solution

Domain: $(0, \infty);$ Range: $(- \infty, \infty);$ Vertical asymptote: $x = 0;$ x-intercept: $(e^{3}, 0);$ y-intercept: DNE

Graphical

For the following exercises, match each function in Figure 17 with the letter corresponding to its graph.

Graph of five logarithmic functions. — Figure 17

Exercise 26

$d (x) = \log (x)$

Exercise 27

$f (x) = \ln (x)$

Solution

B

Exercise 28

$g (x) = \log_{2} (x)$

Exercise 29

$h (x) = \log_{5} (x)$

Solution

C

Exercise 30

$j (x) = \log_{25} (x)$

For the following exercises, match each function in Figure 18 with the letter corresponding to its graph.

Graph of three logarithmic functions. — Figure 18

Exercise 31

$f (x) = \log_{\frac{1}{3}} (x)$

Solution

B

Exercise 32

$g (x) = \log_{2} (x)$

Exercise 33

$h (x) = \log_{\frac{3}{4}} (x)$

Solution

C

For the following exercises, sketch the graphs of each pair of functions on the same axis.

Exercise 34

$f (x) = \log (x)$ and $g (x) = 10^{x}$

Exercise 35

$f (x) = \log (x)$ and $g (x) = \log_{\frac{1}{2}} (x)$

Solution

Graph of two functions, g(x) = log_(1/2)(x) in orange and f(x)=log(x) in blue.

Exercise 36

$f (x) = \log_{4} (x)$ and $g (x) = \ln (x)$

Exercise 37

$f (x) = e^{x}$ and $g (x) = \ln (x)$

Solution

Graph of two functions, g(x) = ln(1/2)(x) in orange and f(x)=e^(x) in blue.

For the following exercises, match each function in Figure 19 with the letter corresponding to its graph.

Exercise 38

$f (x) = \log_{4} (- x + 2)$

Exercise 39

$g (x) = - \log_{4} (x + 2)$

Solution

C

Exercise 40

$h (x) = \log_{4} (x + 2)$

For the following exercises, sketch the graph of the indicated function.

Exercise 41

$f (x) = \log_{2} (x + 2)$

Solution

Exercise 42

$f (x) = 2 \log (x)$

Exercise 43

$f (x) = \ln (- x)$

Solution

Exercise 44

$g (x) = \log (4 x + 16) + 4$

Exercise 45

$g (x) = \log (6 - 3 x) + 1$

Solution

Exercise 46

$h (x) = - \frac{1}{2} \ln (x + 1) - 3$

For the following exercises, write a logarithmic equation corresponding to the graph shown.

Exercise 47

Use $y = \log_{2} (x)$ as the parent function.

The graph y=log_2(x) has been reflected over the y-axis and shifted to the right by 1.

Solution

$f (x) = \log_{2} (- (x - 1))$

Exercise 48

Use $f (x) = \log_{3} (x)$ as the parent function.

The graph y=log_3(x) has been reflected over the x-axis, vertically stretched by 3, and shifted to the left by 4.

Exercise 49

Use $f (x) = \log_{4} (x)$ as the parent function.

The graph y=log_4(x) has been vertically stretched by 3, and shifted to the left by 2.

Solution

$f (x) = 3 \log_{4} (x + 2)$

Exercise 50

Use $f (x) = \log_{5} (x)$ as the parent function.

The graph y=log_3(x) has been reflected over the x-axis and y-axis, vertically stretched by 2, and shifted to the right by 5.

Technology

For the following exercises, use a graphing calculator to find approximate solutions to each equation.

Exercise 51

$\log (x - 1) + 2 = \ln (x - 1) + 2$

Solution

$x = 2$

Exercise 52

$\log (2 x - 3) + 2 = - \log (2 x - 3) + 5$

Exercise 53

$\ln (x - 2) = - \ln (x + 1)$

Solution

$x \approx 2 .303$

Exercise 54

$2 \ln (5 x + 1) = \frac{1}{2} \ln (- 5 x) + 1$

Exercise 55

$\frac{1}{3} \log (1 - x) = \log (x + 1) + \frac{1}{3}$

Solution

$x \approx - 0.472$

Extensions

Exercise 56

Let $b$ be any positive real number such that $b \neq 1.$ What must $\log_{b} 1$ be equal to? Verify the result.

Exercise 57

Explore and discuss the graphs of $f (x) = \log_{\frac{1}{2}} (x)$ and $g (x) = - \log_{2} (x) .$ Make a conjecture based on the result.

Solution

The graphs of $f (x) = \log_{\frac{1}{2}} (x)$ and $g (x) = - \log_{2} (x)$ appear to be the same; Conjecture: for any positive base $b \neq 1,$ $\log_{b} (x) = - \log_{\frac{1}{b}} (x) .$

Exercise 58

Prove the conjecture made in the previous exercise.

Exercise 59

What is the domain of the function $f (x) = \ln (\frac{x + 2}{x - 4}) ?$ Discuss the result.

Solution

Recall that the argument of a logarithmic function must be positive, so we determine where $\frac{x + 2}{x - 4} > 0$ . From the graph of the function $f (x) = \frac{x + 2}{x - 4},$ note that the graph lies above the x-axis on the interval $(- \infty, - 2)$ and again to the right of the vertical asymptote, that is $(4, \infty) .$ Therefore, the domain is $(- \infty, - 2) \cup (4, \infty) .$

A graph of a rational function is shown on an xy-coordinate plane. The x-axis ranges from -10 to 10, and the y-axis ranges from -10 to 10. A vertical dashed orange line, labeled "x = 4", represents a vertical asymptote. The graph consists of two smooth blue curves. The left curve extends from negative infinity in the second quadrant, passes through the x-intercept at (-2, 0) (marked with a blue dot), and decreases towards negative infinity as it approaches the vertical asymptote x=4 from the left. The right curve begins at positive infinity as it approaches the vertical asymptote x=4 from the right, then decreases and approaches the x-axis from above as x tends towards positive infinity in the first quadrant.

Exercise 60

Use properties of exponents to find the x-intercepts of the function $f (x) = \log (x^{2} + 4 x + 4)$ algebraically. Show the steps for solving, and then verify the result by graphing the function.

Graphs of Logarithmic Functions

Learning Objectives

Objective 1: Find the domain and range of a relation and a function. (IA 3.5.1)

Practice Makes Perfect

Objective 2: Graph Logarithmic functions. (IA 10.3.3)

Practice Makes Perfect

Finding the Domain of a Logarithmic Function

Identifying the Domain of a Logarithmic Shift

Identifying the Domain of a Logarithmic Shift and Reflection

Graphing Logarithmic Functions

Graphing a Logarithmic Function with the Form f(x) = logb(x).

Graphing Transformations of Logarithmic Functions

Graphing a Horizontal Shift of f(x) = logb(x)

Graphing a Horizontal Shift of the Parent Function y = logb(x)

Graphing a Vertical Shift of f(x) = logb(x)

Graphing a Vertical Shift of the Parent Function f(x) = logb(x)

Graphing Stretches and Compressions of f(x) = logb(x)

Graphing a Stretch or Compression of the Parent Function f(x) = logb(x)

Combining a Shift and a Stretch

Graphing Reflections of f(x) = logb(x)

Graphing a Reflection of a Logarithmic Function

Approximating the Solution of a Logarithmic Equation

Summarizing Translations of the Logarithmic Function

Finding the Vertical Asymptote of a Logarithm Graph

Analysis

Finding the Equation from a Graph

Analysis

Key Equations

Key Concepts

Section Exercises

Verbal

Algebraic

Graphical

Technology

Extensions

Graphing a Logarithmic Function with the Form f(x) = log_b(x).

Graphing a Horizontal Shift of f(x) = log_b(x)

Graphing a Horizontal Shift of the Parent Function y = log_b(x)

Graphing a Vertical Shift of f(x) = log_b(x)

Graphing a Vertical Shift of the Parent Function f(x) = log_b(x)

Graphing Stretches and Compressions of f(x) = log_b(x)

Graphing a Stretch or Compression of the Parent Function f(x) = log_b(x)

Graphing Reflections of f(x) = log_b(x)