7.09 Graphs of logarithmic functions

Graphs of logarithmic functions

The graph of a logarithmic function $y=\log_ax$y=loga​x is related to the graph of the exponential function $y=a^x$y=ax. In particular, they are a reflection of each other across the line $y=x$y=x. This is because exponential and logarithmic functions are inverse functions.  It may be useful to re-visit exponential graphs and their transformations before working through this section. 

Graphs of $a^x$ax and $\log_ax$loga​x, for $a>1$a>1

As $a^x$ax and $\log(a,x)$log(a,x) are a reflection of each other, we can observe the following properties of the graphs of logarithmic functions of the form $y=\log_bx$y=logb​x:

• Domain: the argument, $x$x, is restricted to only positive values. That is, $x>0$x>0.
• Range: all real $y$y values can be obtained by a logarithm
• Asymptotes: there is a vertical asymptote at $x=0$x=0 (on the $y$y-axis) for all logarithmic functions. As a result, there is no $y$y-intercept.
• $x$x-intercept: The logarithm of $1$1 is $0$0, regardless of the base used. As a result, the graph of a logarithmic function intersects the $x$x-axis at $\left(1,0\right)$(1,0).

If the base $b$b is greater than $1$1 then the function increases across the entire domain $x>0$x>0. For $00<b<1 the function decreases across its domain.

The following graph shows $y=\log_2\left(x\right)$y=log2​(x) and $y=\log_{0.5}\left(x\right)$y=log0.5​(x) illustrating the distinctive shape of the log curve.

Two particular log curves from the family of log functions with $b>1$b>1 are shown below. The green curve is the curve of the function $f(x)=\log_2\left(x\right)$f(x)=log2​(x), and the blue curve is the curve of the function $g(x)=\log_4\left(x\right)$g(x)=log4​(x).

The points shown on each curve help to demonstrate the way the gradient of the curve changes as $b$b increases in value.

The following log graph applet allows you to experiment with different bases. You should note that as the base increases beyond $1$1 the rate of increase in the size of the logarithm decreases.

As you move the base back again closer and closer to $1$1 from above, the rate increases so that the curve becomes more and more vertical. You should be able to see why the function cannot exist for bases equal to $1$1.

For positive bases less than $1$1, try moving the slider across the full range of values. What do you notice?

Practice question

Question 1

Transformations of the logarithmic graph $y=k\log_ax+c$y=kloga​x+c

The logarithmic graph can be transformed in a number of ways. The following summarises and revises transformations for a general function $f(x)$f(x):

These transformations are now explored with specific application to the function $y=k\log_ax+c$y=kloga​x+c.

Vertical translation

Recall that adding a constant to a function corresponds to translating the graph vertically. So the graph of $g\left(x\right)=\log_ax+c$g(x)=loga​x+c is a vertical translation of the graph of $f\left(x\right)=\log_ax$f(x)=loga​x. The translation is upwards if $c$c is positive, and downwards if $c$c is negative.

Graphs of $f\left(x\right)=\log_ax$f(x)=loga​x and $g\left(x\right)=\log_ax+c$g(x)=loga​x+c, for $c<0$c<0.

Notice that the asymptote is not changed by a vertical translation, and is still the line $x=0$x=0. The $x$x-intercept has changed however, and now occurs at a point further along the $x$x-axis. The original $x$x-intercept (which was at $\left(1,0\right)$(1,0)) has now been translated vertically to $\left(1,k\right)$(1,k) and is no longer on the $x$x-axis.

Worked example

example 1

The graphs of the function $f\left(x\right)=\log_3\left(-x\right)$f(x)=log3​(−x) and another function $g\left(x\right)$g(x) are shown below.

(a) Describe the transformation used to get from $f\left(x\right)$f(x) to $g\left(x\right)$g(x).

Think: $g\left(x\right)$g(x) has the same general shape as $f\left(x\right)$f(x), just translated upwards. We can figure out how far it has been translated by looking at the distance between corresponding points.

Do: The point on $g\left(x\right)$g(x) that is directly above the $x$x-intercept of $f\left(x\right)$f(x) is at $\left(-1,5\right)$(−1,5), which is $5$5 units higher. In fact, we can see the constant distance of $5$5 units all the way along the function:

So $f\left(x\right)$f(x) has been translated $5$5 units upwards to give $g\left(x\right)$g(x).

(b) Determine the equation of the function $g\left(x\right)$g(x).

Think: We know that $f\left(x\right)$f(x) has been vertically translated $5$5 units upwards to give $g\left(x\right)$g(x). That is, the function has been increased by $5$5.

Do: This means that $g\left(x\right)=\log_3\left(-x\right)+5$g(x)=log3​(−x)+5. This function has an asymptote at $x=0$x=0, and the negative coefficient of $x$x means that it takes values to the left of the asymptote, just like $f\left(x\right)$f(x).
 

Practice questions

Question 2

Question 3

Dilation

Recall that multiplying a function by a constant corresponds to vertically rescaling the function (making it larger or smaller). The graph of $g\left(x\right)=k\log_ax$g(x)=kloga​x is a vertical dilation of the graph of $f\left(x\right)=\log_ax$f(x)=loga​x if $\left|k\right|$|k| is greater than $1$1, and a vertical compression if $\left|k\right|$|k| is between $0$0 and $1$1. Recall that $\left|k\right|$|k| is the absolute value of $k$k, and is the non-negative value of $k$k without regard to its sign.

Graphs of $f\left(x\right)=\log_ax$f(x)=loga​x and $g\left(x\right)=k\log_ax$g(x)=kloga​x, for $00<k<1.

Additionally, if the coefficient $k$k is negative there is also a reflection across the $x$x-axis.

Graphs of $f\left(x\right)=\log_ax$f(x)=loga​x and $g\left(x\right)=k\log_ax$g(x)=kloga​x, for $k<-1$k<−1.

Notice that the asymptote is not changed by this type of transformation, and is still the line $x=0$x=0. The $x$x-intercept also remains unchanged, since multiplying a $y$y-coordinate of $0$0 by any constant $k$k results in $0$0.

Every other point on the graph, however, moves further away from the $x$x-axis (if $\left|k\right|>1$|k|>1) or closer to the $x$x-axis (if $0<\left|k\right|<1$0<|k|<1).

Let's look at an example involving a horizontal reflection too.

Worked example

Example 2

The graphs of the function $f\left(x\right)=\log_4\left(-x\right)$f(x)=log4​(−x) and another function $g\left(x\right)$g(x) are shown below.

Determine the equation of the function $g\left(x\right)$g(x).

Think: $g\left(x\right)$g(x) is upside-down relative to $f\left(x\right)$f(x), and is stretched so that its corresponding points are further away from the $x$x-axis. So there has been a vertical dilation and a reflection about the $x$x-axis. This means that $g\left(x\right)$g(x) will be of the form $g\left(x\right)=a\log_4\left(-x\right)$g(x)=alog4​(−x) where $a<-1$a<−1.

Do: To determine the particular dilation, let's look at the point $\left(-4,1\right)$(−4,1) on the graph of $f\left(x\right)$f(x). The corresponding point on the graph of $g\left(x\right)$g(x) is $\left(-4,-3\right)$(−4,−3).

To get from a $y$y-value of $1$1 to a $y$y-value of $-3$−3, we have multiplied by $-3$−3. So the value of $a$a must be $-3$−3, and therefore the function is $g\left(x\right)=-3\log_4\left(-x\right)$g(x)=−3log4​(−x).

Practice question

Question 4