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5 Nonlinear relationships
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10&10a

5.06 Logarithmic graphs and scales

Lesson
Worksheet
Practice
Lesson

Features of logarithmic graphs

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Graphs of logarithmic equations of the form \\ y=a\log_{B}(x-h) + k (where a,\,B, and h and k are any number and B>0) are called logarithmic graphs.

This is the logarithmic graph of y=\log_{2}x.

Like lines, logarithmic graphs will always have an x-intercept. This is the point on the graph which touches the x-axis. We can find this by setting y=0 and finding the value of x. For example, the x-intercept of y=\log_{2}x is (1,0).

Similarly, we can look for y-intercepts by setting x=0 and then solving for y. Because this is a logarithmic equation, there could be 0 or 1 solutions, and there will be the same number of y-intercepts. For example, the graph of y=\log_{2}x has no y-intercept.

Logarithmic graphs have a vertical asymptote which is the vertical line which the graph approaches but does not touch. For example, the vertical asymptote of y=\log_{2}x is x=0.

Examples

Example 1

Consider the function y = \log_{4} x.

a

Complete the table of values for y = \log_{4} x, rounding any necessary values to two decimal places.

x0.3123451020
y-0.870.791.66
Worked Solution
Create a strategy

Use the change of base law: \log_{a} b = \dfrac{\log_{10} b}{\log_{10} a}

Apply the idea

Substitute each of the x-values in the table into the equation and evaluate.

For x=1:

\displaystyle y\displaystyle =\displaystyle \log_{4} x
\displaystyle =\displaystyle \log_{4} 1Substitute x=1
\displaystyle =\displaystyle \dfrac{\log_{10} 1}{\log_{10} 4}Use the change of base law
\displaystyle =\displaystyle 0Evaluate

Similarly, by substituting the remaining x-values into \log_{4} x, we get:

x0.3123451020
y-0.8700.50.7911.161.662.16
b

Which of the following is the graph of y = \log_{4} x?

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Worked Solution
Create a strategy

Plot some of the points from the table of values and draw the curve passing through each plotted point.

Apply the idea
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Looking at the table of values, notice that as x increases, y increases. So we expect to have an increasing function.

Some of the ordered pairs of points to be plotted on the coordinate plane are (0.3,-0.87),\, (1,0),\, (2,0.5), and (4,1).

We can draw the curve passing through each of these plotted points, so the correct answer is option B.

Idea summary

The graph of a logarithmic equation of the form y=a\log_{B}(x-h) + k is a logarithmic graph.

Logarithmic graphs have an x-intercept and can have 0 or 1 y-intercepts, depending on the solutions to the logarithmic equation.

Logarithmic graphs have a vertical asymptote which is the vertical line that the graph approaches but does not intersect.

Transformations of logarithmic graphs

A logarithmic graph can be vertically translated by increasing or decreasing the y-values by a constant number. So to translate y=\log_{2}x up by k units gives us y=\log_{2}x + k.

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This graph shows y=\log_{2}x translated vertically up by 2 to get y=\log_{2}x + 2, and down by 2 to get y=\log_{2}x -2.

Similarly, a logarithmic graphh can be horizontally translated by increasing or decreasing the x-values by a constant number. However, the x-value together with the translation must both be in the logarithm. That is, to translate y=\log_{2}x to the left by h units we get y=\log_{2}(x+h).

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This graph shows y=\log_{2}x translated horizontally to the left by 2 to get y=\log_{2}(x+2), and to the right by 2 to get y=\log_{2}(x-2).

A logarithmic graph can be vertically scaled by multiplying every y-value by a constant number. So to expand the logarithmic graph y=\log_{2}x by a scale factor of a we get y=a \log_{2}x. We can compress an exponential graph by dividing by the scale factor instead.

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This graph shows y=\log_{2}x vertically expanded by a scale factor of 2 to get y=2\log_{2}x and compressed by a scale factor of 2 to gety=\dfrac{1}{2}\log_{2}x.

We can vertically reflect a logarithmic graph about the x-axis by taking the negative of the y-values. So to reflect y=\log_{2}x about the x-axis gives us y=-\log_{2}x.

We can similarly horizontally reflect a logarithmic graph about the y-axis by taking the negative of the x-values. So to reflect y=\log_{2}x about the y-axis gives us y=\log_{2}(-x).

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This graph shows y=\log_{2}x horizontally reflected or graph about the y-axis to get y=-\log_{2}x and vertically reflected or graph about the x-axis to get y=\log_{2}(-x).

Exploration

Use the following applet to explore transformations of the graph of a logarithmic function by dragging the sliders.

Loading interactive...

Changing B changes the steepness of the graph. Changing A changes the steepness of the graph and negative values of A flip the curve horizontally. Changing h shifts the curve horizontally, and changing k shifts the curve vertically.

Examples

Example 2

A graph of the function y = \log_{3} x is shown below.

A graph of the function y = \log_{3} x + 3 can be obtained from the original graph by transforming it in some way.

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a

Complete the table of values below for y=\log_{3} x:

x\dfrac{1}{3}139
\log_3 x
Worked Solution
Create a strategy

Substitute each of the x-values in the table into the equation and evaluate.

Apply the idea

For x=\dfrac{1}{3}=0.3333:

\displaystyle y\displaystyle =\displaystyle \log_{3} x
\displaystyle =\displaystyle \log_{3} 0.3333Substitute x=0.3333
\displaystyle =\displaystyle \dfrac{\log_{10} 0.3333}{\log_{10} 3}Use the change of base law
\displaystyle =\displaystyle -1Evaluate

Similarly, by substituting the remaining x-values into \log_{3} x, we get:

x\dfrac{1}{3}139
\log_3 x-1012
b

Now complete the table of values below for y=\log_{3} x + 3:

x\dfrac{1}{3}139
\log_3 x +3
Worked Solution
Create a strategy

Add 3 to each of the resulting y-values of the table for y=\log_{3} x.

Apply the idea

In part (a) we have the tables of values for y=\log_{3} x, such as if x=\dfrac{1}{3}, \log_{3} x=-1.

x\dfrac{1}{3}139
\log_3 x-1012

So for y=\log_{3} x + 3, \, x=\dfrac{1}{3} we have:

\displaystyle y\displaystyle =\displaystyle \log_{3} x + 3
\displaystyle =\displaystyle \log_{3} \dfrac{1}{3} + 3Substitute x=\dfrac{1}{3}
\displaystyle =\displaystyle -1+3Substitute \log_{3} \dfrac{1}{3}=-1
\displaystyle =\displaystyle 2Evaluate

Similarly, by substituting the remaining x-values into \log_{3} x +3, we get:

x\dfrac{1}{3}139
\log_3 x2345
c

Which of the following is a graph of y=\log_{3} x +3?

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Worked Solution
Create a strategy

Plot the points in the table of values and draw the curve passing through each plotted point.

Apply the idea
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From the table of values in part (b) we have some the ordered pairs of points to be plotted on the coordinate plane: (1,3), (3,4) , and (9,5).

This curve of the equation \log_{4} x + 3 must pass through each of the plotted points.

So option B is the correct answer.

d

Which features of the graph are unchanged after it has been translated 3 units upwards?

A
The range.
B
The general shape of the graph.
C
The vertical asymptote.
D
The x-intercept.
Worked Solution
Create a strategy

Compare the graphs of the two equations.

Apply the idea
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Comparing the two graphs of the wo equations y=\log_{3} and y=\log_{3} x +3, we can see from the graph that the range, the general shape of the graph and the vertical asypmtote are unchanged after it has been translated 3 units upward.

So options A, B and C are the correct answers.

Example 3

Given the graph of y=\log_{6} (-x) , draw the graph of y=5\log_{6} (-x) on the same plane.

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Worked Solution
Create a strategy

Vertically expand the given graph by a scale factor 5.

Apply the idea
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The graph of y=5 \log_{6} (-x) can be obtained from the graph of y=5 \log_{6} (-x) by vertically expanding it with a scale factor 5.

The transformation does not change the location of the asymptote or the x-intercept, we will only need to change the location of the a point on the graph.

Let us look at the point (-6,1) on the original graph. The new y-coordinate of this point when it is multiplied by 5 is (-6,5).

So we have the graph of y=5 \log_{6} (-x).

Idea summary

Logarithmic graphs can be transformed in the following ways (starting with the logarithmic graph defined by y=\log_{2} x:

  • Vertically translated by k units: y=\log_{2} x + k

  • Horizontally translated by h units: y=\log_{2} (x-h)

  • Vertically scaled by a scale factor of a: y=a\log_{2} x

  • Vertically reflected about the x-axis: y=-\log_{2} x

  • Horizontally reflected about the y-axis: y=\log_{2} (-x)

Outcomes

ACMNA265 (10a)

Use the definition of a logarithm to establish and apply the laws of logarithms

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