The graph of a logarithmic function $y=\log_ax$y=logax is unsurprisingly 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.
As these functions 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=logbx:
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 red 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 shown 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.
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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?
Consider the function $y=\log_4x$y=log4x, the graph of which has been sketched below.
Complete the following table of values.
$x$x | $\frac{1}{16}$116 | $\frac{1}{4}$14 | $4$4 | $16$16 | $256$256 |
$y$y | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Determine the $x$x-value of the $x$x-intercept of $y=\log_4x$y=log4x.
How many $y$y-intercepts does $\log_4x$log4x have?
Determine the $x$x value for which $\log_4x=1$log4x=1.
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)=logax+c is a vertical translation of the graph of $f\left(x\right)=\log_ax$f(x)=logax. The translation is upwards if $c$c is positive, and downwards if $c$c is negative.
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.
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-intercepts 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).
A function of the form $y=k\log_ax+c$y=klogax+c represents a vertical translation by $c$c units of the function $y=\log_ax$y=logax.
Which of the following options shows the graph of $y=\log_3x$y=log3x after it has been translated $2$2 units up?
The function $y=\log_5x$y=log5x is translated downwards by $2$2 units.
State the equation of the function after it has been translated.
The graph of $y=\log_5x$y=log5x is shown below. Draw the translated graph on the same plane.
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)=klogax is a vertical dilation of the graph of $f\left(x\right)=\log_ax$f(x)=logax 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.
Additionally, if the coefficient $k$k is negative there is also a reflection across the $x$x-axis.
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.
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 across 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).
A function of the form $y=k\log_ax$y=klogax represents a vertical rescaling of the function $y=\log_ax$y=logax.
The graph of $y=\log_7x$y=log7x is shown below.
What transformation of the graph of $y=\log_7x$y=log7x is needed to get the graph of $y=-3\log_7x$y=−3log7x?
Reflection across the $x$x-axis only.
Vertical compression by a factor of $3$3 and reflection across the $x$x-axis.
Vertical dilation by a factor of $3$3 and reflection across the $x$x-axis.
Vertical dilation by a factor of $3$3 only.
Vertical compression by a factor of $3$3 only.
Now draw the graph of $y=-3\log_7x$y=−3log7x on the same plane as $y=\log_7x$y=log7x: