Multiple transformations of log(x)

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=logb​x. 

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=logb​x 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=−3log2​x and $y=\log_2x$y=log2​x. 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=logb​x(mx+c) allows for the repositioning of the domain.

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

For the function $y=\log_bx\left(mx+c\right)$y=logb​x(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=logb​x 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=log2​x with $y=\log_2\left(3x-4\right)$y=log2​(3x−4). 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=log2​x with $y=\log_2x+3$y=log2​x+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 $\log_2x+3=0$log2​x+3=0. This means $\log_2x=-3$log2​x=−3, or that $x=2^{-3}=\frac{1}{8}$x=2−3=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​(2x−6)+4. It can be thought of as the basic function $y=\log_2x$y=log2​x 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​(2x−6)+4=0 as follows:

$-3\log_2\left(2x-6\right)+4$−3log2​(2x−6)+4	$=$=	$0$0
$3\log_2\left(2x-6\right)$3log2​(2x−6)	$=$=	$4$4
$\log_2\left(2x-6\right)$log2​(2x−6)	$=$=	$\frac{4}{3}$43​
$2x-6$2x−6	$=$=	$2^{\frac{4}{3}}$243​
$2x$2x	$=$=	$8.5198421$8.5198421
$x$x	$=$=	$4.26$4.26

The domain is found by solving $2x-6>0$2x−6>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​(x−h)+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

Question 2