6.05 Graphs of logarithms

Graphs of logarithmic functions

For any given base, $b$b, the graph of a logarithmic function $y=\log_bx$y=logb​x is related to the graph of the exponential function $y=b^x$y=bx. 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 $b^x$bx and $\log_bx$logb​x, for $b>1$b>1

As $b^x$bx and $\log_bx$logb​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.
• Asymptotes: there is a vertical asymptote at $x=0$x=0 (on the $y$y-axis) for any logarithmic function of the form $y=\log_bx$y=logb​x, regardless of the base. As a result, there is no $y$y-intercept.
• $x$x-intercept: The logarithm of $1$1 is $0$0, irrespective of the base used. As a result, the graph of $y=\log_bx$y=logb​x intersects the $x$x-axis at $\left(1,0\right)$(1,0).

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

The points shown on each curve help to demonstrate the way the gradient of the curve changes as $b$b increases in value. The larger the value of the base the less steep the graph is. Each graph will pass through $\left(b,1\right)$(b,1), since the $\log_bb=1$logb​b=1. Thus, the higher base graph will appear closer to the $x$x-axis.

Just as for exponential functions if the base $b$b is greater than $1$1 then the function increases across the entire domain and for $00<b<1 the function decreases across its domain. This course will focus on logarithmic functions with bases greater than $1$1.

Practice question

Question 1

Question 2

Transformations of the logarithmic graph $y=\log_bx$y=logb​x

The logarithmic graph can be transformed in a number of ways. The following summarises and revises these transformations:

Translations

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

Graphs of $f\left(x\right)=\log_bx$f(x)=logb​x and $g\left(x\right)=\log_bx+k$g(x)=logb​x+k, for $k<0$k<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, this can be found by solving for $x$x when $y=0$y=0. 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).

Adding a constant to the argument of a logarithmic function corresponds to translating the graph horizontally. So the graph of $g\left(x\right)=\log_b\left(x-h\right)$g(x)=logb​(x−h) is a horizontal translation of the graph of $f\left(x\right)=\log_bx$f(x)=logb​x. The translation is to the right if $h$h is positive, and to the left if $h$h is negative.

Graphs of $f\left(x\right)=\log_b\left(x\right)$f(x)=logb​(x) and $g\left(x\right)=\log_b\left(x-h\right)$g(x)=logb​(x−h), for $h>0$h>0.

Notice that the asymptote is changed by a horizontal translation, and is translated to the line $x=h$x=h. The $x$x-intercept has also changed, it has been translated horizontally by $h$h units and now occurs at $\left(1+h,0\right)$(1+h,0).

Worked example

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.

Practice questions

Question 3

Question 4

Question 5

Question 6

Vertical dilation

Recall that multiplying a function by a constant corresponds to vertically rescaling the function (stretching) from the horizontal axis. The graph of $g\left(x\right)=a\log_bx$g(x)=alogb​x is a vertical dilation of the graph of $f\left(x\right)=\log_bx$f(x)=logb​x if $\left|a\right|$|a| is greater than $1$1, and a vertical compression if $\left|a\right|$|a| is between $0$0 and $1$1.

Graphs of $f\left(x\right)=\log_bx$f(x)=logb​x and $g\left(x\right)=a\log_bx$g(x)=alogb​x, for $00<a<1.

Does this graph look familiar? How does it compare to the graph near the start of the lesson comparing logarithmic functions with different bases? In fact, we can consider a change of base as a vertical dilation. We can show the equivalence using the change of base rule, which states:

$\log_b(x)=\frac{\log_d(x)}{\log_d(b)}$logb​(x)=logd​(x)logd​(b)​

That is, if we compare the graph of $f\left(x\right)=\log_d\left(x\right)$f(x)=logd​(x) to $g\left(x\right)=\log_b\left(x\right)$g(x)=logb​(x), $g(x)$g(x) will be a dilation of $f(x)$f(x) by a factor of $\frac{1}{\log_d\left(b\right)}$1logd​(b)​ from the $x$x-axis.

For example, if we compare the graphs of $f\left(x\right)=\log_2\left(x\right)$f(x)=log2​(x) to $g\left(x\right)=\log_4\left(x\right)$g(x)=log4​(x), we can see the graph of $g\left(x\right)$g(x) is a dilation of $f\left(x\right)$f(x) by a factor of $\frac{1}{2}=\frac{1}{\log_24}$12​=1log2​4​.

Reflection

If the coefficient $a$a is negative this will result in a reflection across the $x$x-axis.

Graphs of $f\left(x\right)=\log_2x$f(x)=log2​x and $g\left(x\right)=-\log_2x$g(x)=−log2​x, example of $a=-1$a=−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 $a$a results in $0$0.

Horizontal dilation

Horizontal dilations are not a focus of this course, but we will briefly consider them. From our previous work of transformations of general functions, the graph of $g\left(x\right)=\log_b\left(cx\right)$g(x)=logb​(cx) can be obtained from the graph of $f\left(x\right)=\log_bx$f(x)=logb​x as follows:

• $c$c dilates (stretches) the graph by a factor of $\frac{1}{c}$1c​ from the $y$y-axis, parallel to the $x$x-axis

• When $c<0$c<0 the graph is reflected across the $y$y-axis

However, with our knowledge of logarithm laws, we can treat this dilation factor as a vertical shift.

For example, $g\left(x\right)=\log_3\left(9x\right)$g(x)=log3​(9x) can be rearranged using our log laws as follows:

Hence, the graph of $g\left(x\right)$g(x) is equivalent to a vertical translation of $2$2 units upwards of the graph of $f(x)=\log_3\left(x\right)$f(x)=log3​(x).

Practice question

Question 7

Question 8