Graphs of Logarithmic Functions

Learning Objectives

In this section, you will:

• Identify the domain of a logarithmic function.
• Graph logarithmic functions.

In Graphs of Exponential Functions, we saw how creating a graphical representation of an exponential model gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the cause for an effect.

To illustrate, suppose we invest $2500$2500 in an account that offers an annual interest rate of 5%,5%, compounded continuously. We already know that the balance in our account for any year t𝑡 can be found with the equation A=2500e0.05t.𝐴=2500𝑒0.05𝑡.

But what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure 1 shows this point on the logarithmic graph.

Figure 1

In this section we will discuss the values for which a logarithmic function is defined, and then turn our attention to graphing the family of logarithmic functions.

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=bx𝑦=𝑏𝑥 for any real number x𝑥 and constant b>0,𝑏>0, b≠1,𝑏≠1, where

• The domain of y𝑦 is (−∞,∞).(−∞,∞).
• The range of y𝑦 is (0,∞).(0,∞).

In the last section we learned that the logarithmic function y=logb(x)𝑦=log𝑏(𝑥) is the inverse of the exponential function y=bx.𝑦=𝑏𝑥. So, as inverse functions:

• The domain of y=logb(x)𝑦=log𝑏(𝑥) is the range of y=bx:𝑦=𝑏𝑥: (0,∞).(0,∞).
• The range of y=logb(x)𝑦=log𝑏(𝑥) is the domain of y=bx:𝑦=𝑏𝑥: (−∞,∞).(−∞,∞).

Transformations of the parent function y=logb(x)𝑦=log𝑏(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=bx.𝑦=𝑏𝑥. Similarly, applying transformations to the parent function y=logb(x)𝑦=log𝑏(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)=log4(2x−3).𝑓(𝑥)=log4(2𝑥−3). This function is defined for any values of x𝑥 such that the argument, in this case 2x−3,2𝑥−3, is greater than zero. To find the domain, we set up an inequality and solve for x:𝑥:

2x−3>02x>3x>1.5Show the argument greater than zero.Add 3.Divide by 2.2𝑥−3>0Show the argument greater than zero.2𝑥>3Add 3.𝑥>1.5Divide by 2.

In interval notation, the domain of f(x)=log4(2x−3)𝑓(𝑥)=log4(2𝑥−3) is (1.5,∞).(1.5,∞).

HOW TO

Given a logarithmic function, identify the domain.

1. Set up an inequality showing the argument greater than zero.
2. Solve for x.𝑥.
3. Write the domain in interval notation.

EXAMPLE 1

Identifying the Domain of a Logarithmic Shift

What is the domain of f(x)=log2(x+3)?𝑓(𝑥)=log2(𝑥+3)?

Solution

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

x+3>0x>−3The input must be positive.Subtract 3.𝑥+3>0The input must be positive.𝑥>−3Subtract 3.

The domain of f(x)=log2(x+3)𝑓(𝑥)=log2(𝑥+3) is (−3,∞).(−3,∞).

TRY IT #1

What is the domain of f(x)=log5(x−2)+1?𝑓(𝑥)=log5(𝑥−2)+1?

EXAMPLE 2

Identifying the Domain of a Logarithmic Shift and Reflection

What is the domain of f(x)=log(5−2x)?𝑓(𝑥)=log(5−2𝑥)?

Solution

The logarithmic function is defined only when the input is positive, so this function is defined when 5–2x>0.5–2𝑥>0. Solving this inequality,

5−2x>0−2x>−5x<52The input must be positive.Subtract 5.Divide by −2and switch the inequality.5−2𝑥>0The input must be positive.−2𝑥>−5Subtract 5.𝑥<52Divide by −2and switch the inequality.

The domain of f(x)=log(5−2x)𝑓(𝑥)=log(5−2𝑥) is (–∞,52).(–∞,52).

TRY IT #2

What is the domain of f(x)=log(x−5)+2?𝑓(𝑥)=log(𝑥−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=logb(x)𝑦=log𝑏(𝑥) along with all its transformations: shifts, stretches, compressions, and reflections.

We begin with the parent function y=logb(x).𝑦=log𝑏(𝑥). Because every logarithmic function of this form is the inverse of an exponential function with the form y=bx,𝑦=𝑏𝑥, their graphs will be reflections of each other across the line y=x.𝑦=𝑥. To illustrate this, we can observe the relationship between the input and output values of y=2x𝑦=2𝑥 and its equivalent x=log2(y)𝑥=log2(𝑦) in Table 1.

x𝑥	−3−3	−2−2	−1−1	00	11	22	33
2x=y2𝑥=𝑦	1818	1414	1212	11	22	44	88
log2(y)=xlog2(𝑦)=𝑥	−3−3	−2−2	−1−1	00	11	22	33

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)=2x𝑓(𝑥)=2𝑥 and g(x)=log2(x).𝑔(𝑥)=log2(𝑥). See Table 2.

f(x)=2x𝑓(𝑥)=2𝑥	(−3,18)(−3,18)	(−2,14)(−2,14)	(−1,12)(−1,12)	(0,1)(0,1)	(1,2)(1,2)	(2,4)(2,4)	(3,8)(3,8)
g(x)=log2(x)𝑔(𝑥)=log2(𝑥)	(18,−3)(18,−3)	(14,−2)(14,−2)	(12,−1)(12,−1)	(1,0)(1,0)	(2,1)(2,1)	(4,2)(4,2)	(8,3)(8,3)

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

Figure 2 Notice that the graphs of f(x)=2x𝑓(𝑥)=2𝑥 and g(x)=log2(x)𝑔(𝑥)=log2(𝑥) are reflections about the line y=x.𝑦=𝑥.

Observe the following from the graph:

• f(x)=2x𝑓(𝑥)=2𝑥 has a y-intercept at (0,1)(0,1) and g(x)=log2(x)𝑔(𝑥)=log2(𝑥) has an x– intercept at (1,0).(1,0).
• The domain of f(x)=2x,𝑓(𝑥)=2𝑥, (−∞,∞),(−∞,∞), is the same as the range of g(x)=log2(x).𝑔(𝑥)=log2(𝑥).
• The range of f(x)=2x,𝑓(𝑥)=2𝑥, (0,∞),(0,∞), is the same as the domain of g(x)=log2(x).𝑔(𝑥)=log2(𝑥).

CHARACTERISTICS OF THE GRAPH OF THE PARENT FUNCTION, f(x)=logb(x):𝑓(𝑥)=log𝑏(𝑥):

For any real number x𝑥 and constant b>0,𝑏>0, b≠1,𝑏≠1, we can see the following characteristics in the graph of f(x)=logb(x):𝑓(𝑥)=log𝑏(𝑥):

• one-to-one function
• vertical asymptote: x=0𝑥=0
• domain: (0,∞)(0,∞)
• range: (−∞,∞)(−∞,∞)
• x-intercept: (1,0)(1,0) and key point (b,1)(𝑏,1)
• y-intercept: none
• increasing if b>1𝑏>1
• decreasing if 0<b<10<𝑏<1

See Figure 3.

Figure 3

Figure 4 shows how changing the base b𝑏 in f(x)=logb(x)𝑓(𝑥)=log𝑏(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)ln(𝑥) has base e≈2.718.)𝑒≈2.718.)

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

HOW TO

Given a logarithmic function with the form f(x)=logb(x),𝑓(𝑥)=log𝑏(𝑥), graph the function.

1. Draw and label the vertical asymptote, x=0.𝑥=0.
2. Plot the x-intercept, (1,0).(1,0).
3. Plot the key point (b,1).(𝑏,1).
4. Draw a smooth curve through the points.
5. State the domain, (0,∞),(0,∞), the range, (−∞,∞),(−∞,∞), and the vertical asymptote, x=0.𝑥=0.

EXAMPLE 3

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

Graph f(x)=log5(x).𝑓(𝑥)=log5(𝑥). State the domain, range, and asymptote.

Solution

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

• Since b=5𝑏=5 is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote x=0,𝑥=0, and the right tail will increase slowly without bound.
• The x-intercept is (1,0).(1,0).
• The key point (5,1)(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).

Figure 5

The domain is (0,∞),(0,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.𝑥=0.

TRY IT #3

Graph f(x)=log15(x).𝑓(𝑥)=log15(𝑥). State the domain, range, and asymptote.

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=logb(x)𝑦=log𝑏(𝑥) without loss of shape.

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

When a constant c𝑐 is added to the input of the parent function f(x)=logb(x),𝑓(𝑥)=𝑙𝑜𝑔𝑏(𝑥), the result is a horizontal shift c𝑐 units in the opposite direction of the sign on c.𝑐. To visualize horizontal shifts, we can observe the general graph of the parent function f(x)=logb(x)𝑓(𝑥)=log𝑏(𝑥) and for c>0𝑐>0 alongside the shift left, g(x)=logb(x+c),𝑔(𝑥)=log𝑏(𝑥+𝑐), and the shift right, h(x)=logb(x−c).ℎ(𝑥)=log𝑏(𝑥−𝑐). See Figure 6.

Figure 6

HORIZONTAL SHIFTS OF THE PARENT FUNCTION f(x)=logb(x)𝑓(𝑥)=log𝑏(𝑥)

For any constant c,𝑐, the function f(x)=logb(x+c)𝑓(𝑥)=log𝑏(𝑥+𝑐)

• shifts the parent function y=logb(x)𝑦=log𝑏(𝑥) left c𝑐 units if c>0.𝑐>0.
• shifts the parent function y=logb(x)𝑦=log𝑏(𝑥) right c𝑐 units if c<0.𝑐<0.
• has the vertical asymptote x=−c.𝑥=−𝑐.
• has domain (−c,∞).(−𝑐,∞).
• has range (−∞,∞).(−∞,∞).

HOW TO

Given a logarithmic function with the form f(x)=logb(x+c),𝑓(𝑥)=log𝑏(𝑥+𝑐), graph the translation.

1. Identify the horizontal shift:
   1. If c>0,𝑐>0, shift the graph of f(x)=logb(x)𝑓(𝑥)=log𝑏(𝑥) left c𝑐 units.
   2. If c<0,𝑐<0, shift the graph of f(x)=logb(x)𝑓(𝑥)=log𝑏(𝑥) right c𝑐 units.
2. Draw the vertical asymptote x=−c.𝑥=−𝑐.
3. Identify three key points from the parent function. Find new coordinates for the shifted functions by subtracting c𝑐 from the x𝑥 coordinate.
4. Label the three points.
5. The Domain is (−c,∞),(−𝑐,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=−c.𝑥=−𝑐.

EXAMPLE 4

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

Sketch the horizontal shift f(x)=log3(x−2)𝑓(𝑥)=log3(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)=log3(x−2),𝑓(𝑥)=log3(𝑥−2), we notice x+(−2)=x–2.𝑥+(−2)=𝑥–2.

Thus c=−2,𝑐=−2, so c<0.𝑐<0. This means we will shift the function f(x)=log3(x)𝑓(𝑥)=log3(𝑥) right 2 units.

The vertical asymptote is x=−(−2)𝑥=−(−2) or x=2.𝑥=2.

Consider the three key points from the parent function, (13,−1),(13,−1), (1,0),(1,0), and (3,1).(3,1).

The new coordinates are found by adding 2 to the x𝑥 coordinates.

Label the points (73,−1),(73,−1), (3,0),(3,0), and (5,1).(5,1).

The domain is (2,∞),(2,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=2.𝑥=2.

Figure 7

TRY IT #4

Sketch a graph of f(x)=log3(x+4)𝑓(𝑥)=log3(𝑥+4) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.

Graphing a Vertical Shift of y = log_b(x)

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

Figure 8

VERTICAL SHIFTS OF THE PARENT FUNCTION y=logb(x)𝑦=log𝑏(𝑥)

For any constant d,𝑑, the function f(x)=logb(x)+d𝑓(𝑥)=log𝑏(𝑥)+𝑑

• shifts the parent function y=logb(x)𝑦=log𝑏(𝑥) up d𝑑 units if d>0.𝑑>0.
• shifts the parent function y=logb(x)𝑦=log𝑏(𝑥) down d𝑑 units if d<0.𝑑<0.
• has the vertical asymptote x=0.𝑥=0.
• has domain (0,∞).(0,∞).
• has range (−∞,∞).(−∞,∞).

HOW TO

Given a logarithmic function with the form f(x)=logb(x)+d,𝑓(𝑥)=log𝑏(𝑥)+𝑑, graph the translation.

1. Identify the vertical shift:
   • If d>0,𝑑>0, shift the graph of f(x)=logb(x)𝑓(𝑥)=log𝑏(𝑥) up d𝑑 units.
   • If d<0,𝑑<0, shift the graph of f(x)=logb(x)𝑓(𝑥)=log𝑏(𝑥) down d𝑑 units.
2. Draw the vertical asymptote x=0.𝑥=0.
3. Identify three key points from the parent function. Find new coordinates for the shifted functions by adding d𝑑 to the y𝑦 coordinate.
4. Label the three points.
5. The domain is (0,∞),(0,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.𝑥=0.

EXAMPLE 5

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

Sketch a graph of f(x)=log3(x)−2𝑓(𝑥)=log3(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)=log3(x)−2,𝑓(𝑥)=log3(𝑥)−2, we will notice d=–2.𝑑=–2. Thus d<0.𝑑<0.

This means we will shift the function f(x)=log3(x)𝑓(𝑥)=log3(𝑥) down 2 units.

The vertical asymptote is x=0.𝑥=0.

Consider the three key points from the parent function, (13,−1),(13,−1), (1,0),(1,0), and (3,1).(3,1).

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

Label the points (13,−3),(13,−3), (1,−2),(1,−2), and (3,−1).(3,−1).

The domain is (0,∞),(0,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.𝑥=0.

Figure 9

The domain is (0,∞),(0,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.𝑥=0.

TRY IT #5

Sketch a graph of f(x)=log2(x)+2𝑓(𝑥)=log2(𝑥)+2 alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.

Graphing Stretches and Compressions of y = log_b(x)

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

Figure 10

VERTICAL STRETCHES AND COMPRESSIONS OF THE PARENT FUNCTION y=logb(x)𝑦=log𝑏(𝑥)

For any constant a>1,𝑎>1, the function f(x)=alogb(x)𝑓(𝑥)=𝑎log𝑏(𝑥)

• stretches the parent function y=logb(x)𝑦=log𝑏(𝑥) vertically by a factor of a𝑎 if a>1.𝑎>1.
• compresses the parent function y=logb(x)𝑦=log𝑏(𝑥) vertically by a factor of a𝑎 if 0<a<1.0<𝑎<1.
• has the vertical asymptote x=0.𝑥=0.
• has the x-intercept (1,0).(1,0).
• has domain (0,∞).(0,∞).
• has range (−∞,∞).(−∞,∞).

HOW TO

Given a logarithmic function with the form f(x)=alogb(x),𝑓(𝑥)=𝑎log𝑏(𝑥), a>0,𝑎>0, graph the translation.

1. Identify the vertical stretch or compressions:
   • If |a|>1,|𝑎|>1, the graph of f(x)=logb(x)𝑓(𝑥)=log𝑏(𝑥) is stretched by a factor of a𝑎 units.
   • If |a|<1,|𝑎|<1, the graph of f(x)=logb(x)𝑓(𝑥)=log𝑏(𝑥) is compressed by a factor of a𝑎 units.
2. Draw the vertical asymptote x=0.𝑥=0.
3. Identify three key points from the parent function. Find new coordinates for the shifted functions by multiplying the y𝑦 coordinates by a.𝑎.
4. Label the three points.
5. The domain is (0,∞),(0,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.𝑥=0.

EXAMPLE 6

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

Sketch a graph of f(x)=2log4(x)𝑓(𝑥)=2log4(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)=2log4(x),𝑓(𝑥)=2log4(𝑥), we will notice a=2.𝑎=2.

This means we will stretch the function f(x)=log4(x)𝑓(𝑥)=log4(𝑥) by a factor of 2.

The vertical asymptote is x=0.𝑥=0.

Consider the three key points from the parent function, (14,−1),(14,−1), (1,0),(1,0), and (4,1).(4,1).

The new coordinates are found by multiplying the y𝑦 coordinates by 2.

Label the points (14,−2),(14,−2), (1,0),(1,0), and (4,2).(4,2).

The domain is (0,∞),(0,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.𝑥=0. See Figure 11.

Figure 11

The domain is (0,∞),(0,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.𝑥=0.

TRY IT #6

Sketch a graph of f(x)=12log4(x)𝑓(𝑥)=12log4(𝑥) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.

EXAMPLE 7

Combining a Shift and a Stretch

Sketch a graph of f(x)=5log(x+2).𝑓(𝑥)=5log(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.𝑥=−2. The x-intercept will be (−1,0).(−1,0). The domain will be (−2,∞).(−2,∞). Two points will help give the shape of the graph: (−1,0)(−1,0) and (8,5).(8,5). We chose x=8𝑥=8 as the x-coordinate of one point to graph because when x=8,𝑥=8, x+2=10,𝑥+2=10, the base of the common logarithm.

Figure 12

The domain is (−2,∞),(−2,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=−2.𝑥=−2.

TRY IT #7

Sketch a graph of the function f(x)=3log(x−2)+1.𝑓(𝑥)=3log(𝑥−2)+1. State the domain, range, and asymptote.

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

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

Figure 13

REFLECTIONS OF THE PARENT FUNCTION y=logb(x)𝑦=log𝑏(𝑥)

The function f(x)=−logb(x)𝑓(𝑥)=−log𝑏(𝑥)

• reflects the parent function y=logb(x)𝑦=log𝑏(𝑥) about the x-axis.
• has domain, (0,∞),(0,∞), range, (−∞,∞),(−∞,∞), and vertical asymptote, x=0,𝑥=0, which are unchanged from the parent function.

The function f(x)=logb(−x)𝑓(𝑥)=log𝑏(−𝑥)

• reflects the parent function y=logb(x)𝑦=log𝑏(𝑥) about the y-axis.
• has domain (−∞,0).(−∞,0).
• has range, (−∞,∞),(−∞,∞), and vertical asymptote, x=0,𝑥=0, which are unchanged from the parent function.

HOW TO

Given a logarithmic function with the parent function f(x)=logb(x),𝑓(𝑥)=log𝑏(𝑥), graph a translation.

If f(x)=−logb(x)If 𝑓(𝑥)=−log𝑏(𝑥)	If f(x)=logb(−x)If 𝑓(𝑥)=log𝑏(−𝑥)
1. Draw the vertical asymptote, x=0.𝑥=0.	1. Draw the vertical asymptote, x=0.𝑥=0.
2. Plot the x-intercept, (1,0).(1,0).	2. Plot the x-intercept, (1,0).(1,0).
3. Reflect the graph of the parent function f(x)=logb(x)𝑓(𝑥)=log𝑏(𝑥) about the x-axis.	3. Reflect the graph of the parent function f(x)=logb(x)𝑓(𝑥)=log𝑏(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=0𝑥=0.	5. State the domain, (−∞, 0) the range, (−∞, ∞) and the vertical asymptote x=0.𝑥=0.

EXAMPLE 8

Graphing a Reflection of a Logarithmic Function

Sketch a graph of f(x)=log(−x)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),𝑓(𝑥)=log(−𝑥), identify the behavior and key points for the graph.

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

Figure 14

The domain is (−∞,0),(−∞,0), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.𝑥=0.

TRY IT #8

Graph f(x)=−log(−x).𝑓(𝑥)=−log(−𝑥). State the domain, range, and asymptote.

HOW TO

Given a logarithmic equation, use a graphing calculator to approximate solutions.

1. Press [Y=]. Enter the given logarithm equation or equations as Y₁= and, if needed, Y₂=.
2. Press [GRAPH] to observe the graphs of the curves and use [WINDOW] to find an appropriate view of the graphs, including their point(s) of intersection.
3. To find the value of x,𝑥, we compute the point of intersection. Press [2ND] then [CALC]. Select “intersect” and press [ENTER] three times. The point of intersection gives the value of x,𝑥, for the point(s) of intersection.

EXAMPLE 9

Approximating the Solution of a Logarithmic Equation

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

Solution

Press [Y=] and enter 4ln(x)+14ln(𝑥)+1 next to Y₁=. Then enter −2ln(x−1)−2ln(𝑥−1) next to Y₂=. For a window, use the values 0 to 5 for x𝑥 and –10 to 10 for y.𝑦. Press [GRAPH]. The graphs should intersect somewhere a little to right of x=1.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≈1.339.𝑥≈1.339.

TRY IT #9

Solve 5log(x+2)=4−log(x)5log(𝑥+2)=4−log(𝑥) graphically. Round to the nearest thousandth.

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.

Transformations of the Parent Function y=logb(x)𝑦=log𝑏(𝑥)
Transformation	Form
ShiftHorizontally c𝑐 units to the leftVertically d𝑑 units up	y=logb(x+c)+d𝑦=log𝑏(𝑥+𝑐)+𝑑
Stretch and CompressStretch if |a|>1|𝑎|>1Compression if |a|<1|𝑎|<1	y=alogb(x)𝑦=𝑎log𝑏(𝑥)
Reflect about the x-axis	y=−logb(x)𝑦=−log𝑏(𝑥)
Reflect about the y-axis	y=logb(−x)𝑦=log𝑏(−𝑥)
General equation for all translations	y=alogb(x+c)+d𝑦=𝑎log𝑏(𝑥+𝑐)+𝑑

TRANSLATIONS OF LOGARITHMIC FUNCTIONS

All translations of the parent logarithmic function, y=logb(x),𝑦=log𝑏(𝑥), have the form

 f(x)=alogb(x+c)+d 𝑓(𝑥)=𝑎log𝑏(𝑥+𝑐)+𝑑

where the parent function, y=logb(x),b>1,𝑦=log𝑏(𝑥),𝑏>1, is

• shifted vertically up d𝑑 units.
• shifted horizontally to the left c𝑐 units.
• stretched vertically by a factor of |a||𝑎| if |a|>0.|𝑎|>0.
• compressed vertically by a factor of |a||𝑎| if 0<|a|<1.0<|𝑎|<1.
• reflected about the x-axis when a<0.𝑎<0.

For f(x)=log(−x),𝑓(𝑥)=log(−𝑥), the graph of the parent function is reflected about the y-axis.

EXAMPLE 10

Finding the Vertical Asymptote of a Logarithm Graph

What is the vertical asymptote of f(x)=−2log3(x+4)+5?𝑓(𝑥)=−2log3(𝑥+4)+5?

Solution

The vertical asymptote is at x=−4.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.𝑥=−4.

TRY IT #10

What is the vertical asymptote of f(x)=3+ln(x−1)?𝑓(𝑥)=3+ln(𝑥−1)?

EXAMPLE 11

Finding the Equation from a Graph

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

Figure 15

Solution

This graph has a vertical asymptote at x=–2𝑥=–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)=−alog(x+2)+k𝑓(𝑥)=−𝑎log(𝑥+2)+𝑘

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

1=−alog(−1+2)+k1=−alog(1)+k1=kSubstitute (−1,1).Arithmetic.log(1)=0.1=−𝑎log(−1+2)+𝑘Substitute (−1,1).1=−𝑎log(1)+𝑘Arithmetic.1=𝑘log(1)=0.

Next, substituting in (2,–1)(2,–1) ,

−1=−alog(2+2)+1−2=−alog(4) a=2log(4)Plug in (2,−1).Arithmetic.Solve for a.−1=−𝑎log(2+2)+1Plug in (2,−1).−2=−𝑎log(4)Arithmetic. 𝑎=2log(4)Solve for 𝑎.

This gives us the equation f(x)=–2log(4)log(x+2)+1.𝑓(𝑥)=–2log(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.

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

TRY IT #11

Give the equation of the natural logarithm graphed in Figure 16.

Figure 16

Q&A

Is it possible to tell the domain and range and describe the end behavior of a function just by looking at the graph?

Yes, if we know the function is a general logarithmic function. For example, look at the graph in Figure 16. The graph approaches x=−3𝑥=−3 (or thereabouts) more and more closely, so x=−3𝑥=−3 is, or is very close to, the vertical asymptote. It approaches from the right, so the domain is all points to the right, {x|x>−3}.{𝑥|𝑥>−3}. The range, as with all general logarithmic functions, is all real numbers. And we can see the end behavior because the graph goes down as it goes left and up as it goes right. The end behavior is that as x→−3+,f(x)→−∞𝑥→−3+,𝑓(𝑥)→−∞ and as x→∞,f(x)→∞.