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Transformations of Logarithmic graphs


The general function given by $y=a\log_b\left(mx+c\right)+d$y=alogb(mx+c)+d can be construed as a transformation of the  basic log function $y=\log_bx$y=logbx

We will examine the effect of the four introduced constants using specific examples. All of the example graphs below use base 2 logarithms, however the same principles would apply for any base.

The dilation factor 

The factor $a$a is called a dilation factor. Just like other functions with a dilation factor, each ordinate of the basic function $y=\log_bx$y=logbx is either increased or decreased according to the size of this factor. The basic function is reflected across the $x$x-axis If $a$a is negative.  

The graph shown here compares $y=-3\log_2x$y=3log2x and $y=\log_2x$y=log2x. Note the dilation and reflection involved.

The domain CHANGE

The argument $\left(mx+c\right)$(mx+c) in the transformed function $y=\log_bx\left(mx+c\right)$y=logbx(mx+c) allows for the repositioning of the domain.

For the function $y=\log_bx$y=logbx, the domain is given by $x>0$x>0.

For the function $y=\log_bx\left(mx+c\right)$y=logbx(mx+c), the domain is found by setting $mx+c>0$mx+c>0 and the solution of this inequality becomes $x>-\frac{c}{m}$x>cm

An important point needs to be made here. If $m=1$m=1 the transformed curve becomes $y=\log_b\left(x+c\right)$y=logb(x+c). This curve is essentially the same as  $y=\log_bx$y=logbx but shifted horizontally a distance of $-c$c units.  However, when $m$m is not $1$1, the transformation becomes more difficult to describe. We can illustrate this with an example. 

The graph below compares $y=\log_2x$y=log2x with $y=\log_2\left(3x-4\right)$y=log2(3x4). Note that the blue curve, with domain $x>\frac{4}{3}$x>43, has shifted to the right, but not as a simple horizontal translation.  

The vertical translation

The final consideration is the effect of the constant  $d$d. This simply shifts the basic function vertically by an amount $d$d. This type of translation is a uniform one.

The following example compares the function $y=\log_2x$y=log2x with $y=\log_2x+3$y=log2x+3. Be careful to note that this is not the same as $y=\log_2\left(2x+3\right)$y=log2(2x+3)

Shifting the curve upward by $3$3 units means that the $x$x-intercept has shifted to the left. We can work out the extent of shift by setting $\log_2x+3=0$log2x+3=0. This means $\log_2x=-3$log2x=3, or that $x=2^{-3}=\frac{1}{8}$x=23=18

Putting it all together

As an example that puts all of these effects into action, consider the new function given by $y=-3\log_2\left(2x-6\right)+4$y=3log2(2x6)+4. It can be thought of as the basic function $y=\log_2x$y=log2x shifted to the right by $\frac{6}{2}=3$62=3 units, then reflected and dilated by a factor of $3$3, and then finally lifted vertically upwards by $4$4 units. 

The approximate value of the new $x$x intercept is found by solving $-3\log_2\left(2x-6\right)+4=0$3log2(2x6)+4=0 as follows:

$-3\log_2\left(2x-6\right)+4$3log2(2x6)+4 $=$= $0$0
$3\log_2\left(2x-6\right)$3log2(2x6) $=$= $4$4
$\log_2\left(2x-6\right)$log2(2x6) $=$= $\frac{4}{3}$43
$2x-6$2x6 $=$= $2^{\frac{4}{3}}$243
$2x$2x $=$= $8.5198421$8.5198421
$x$x $=$=


The domain is found by solving $2x-6>0$2x6>0. Thus the domain is given by $x>3$x>3, and this implies that the curve does not have a $y$y-intercept. Here is the graph:

The Applet

It is important to experiment with the different effects yourself. The following applet allows you to vary the constants of the log function given by $y=a\log_b\left(x-h\right)+k$y=alogb(xh)+k. Focus on varying each constant separately at first, and then try combinations of constants. See what you can discover yourself.

Worked Examples

Question 1

The graph of $y=\log_{10}x$y=log10x is displayed here.

Loading Graph...

  1. What is the $x$x-intercept of this graph?

    $x$x = $\editable{}$

  2. Does the graph cut the $y$y-axis?




  3. What is the graph's domain?



    All real $x$x





  4. Find the value of $y$y when $x=3$x=3. Write your answer to 2 decimal places.

  5. Find the value of $x$x when $y=3$y=3.

Question 2

The function $y=\log x$y=logx has been transformed into the function $y=5\log\left(x+4\right)-2$y=5log(x+4)2.

Complete the following:

  1. The horizontal translation is $\editable{}$ units to the left.

  2. The vertical translation is $\editable{}$ units down.

  3. The vertical dilation factor is $\editable{}$.



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