Logarithms

Hong Kong

Stage 4 - Stage 5

Lesson

The general function given by $y=a\log_b\left(mx+c\right)+d$`y`=`a``l``o``g``b`(`m``x`+`c`)+`d` can be construed as a transformation of the basic log function $y=\log_bx$`y`=`l``o``g``b``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 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`=`l``o``g``b``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`=−3`l``o``g`2`x` and $y=\log_2x$`y`=`l``o``g`2`x`. Note the dilation and reflection involved.

The *argument* $\left(mx+c\right)$(`m``x`+`c`) in the transformed function $y=\log_bx\left(mx+c\right)$`y`=`l``o``g``b``x`(`m``x`+`c`) allows for the repositioning of the domain.

For the function $y=\log_bx$`y`=`l``o``g``b``x`, the domain is given by $x>0$`x`>0.

For the function $y=\log_bx\left(mx+c\right)$`y`=`l``o``g``b``x`(`m``x`+`c`), the domain is found by setting $mx+c>0$`m``x`+`c`>0 and the solution of this inequality becomes $x>-\frac{c}{m}$`x`>−`c``m`.

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`=`l``o``g``b`(`x`+`c`). This curve is essentially the same as $y=\log_bx$`y`=`l``o``g``b``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`=`l``o``g`2`x` with $y=\log_2\left(3x-4\right)$`y`=`l``o``g`2(3`x`−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 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`=`l``o``g`2`x` with $y=\log_2x+3$`y`=`l``o``g`2`x`+3. Be careful to note that this is not the same as $y=\log_2\left(2x+3\right)$`y`=`l``o``g`2(2`x`+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 $\log_2x+3=0$`l``o``g`2`x`+3=0. This means $\log_2x=-3$`l``o``g`2`x`=−3, or that $x=2^{-3}=\frac{1}{8}$`x`=2−3=18.

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`=−3`l``o``g`2(2`x`−6)+4. It can be thought of as the basic function $y=\log_2x$`y`=`l``o``g`2`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 $-3\log_2\left(2x-6\right)+4=0$−3`l``o``g`2(2`x`−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$2`x`−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:

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`=`a``l``o``g``b`(`x`−`h`)+`k`. Focus on varying each constant separately at first, and then try combinations of constants. See what you can discover yourself.

The graph of $y=\log_{10}x$`y`=`l``o``g`10`x` is displayed here.

Loading Graph...

What is the $x$

`x`-intercept of this graph?$x$

`x`= $\editable{}$Does the graph cut the $y$

`y`-axis?Yes

ANo

BWhat is the graph's domain?

$x\ge0$

`x`≥0AAll real $x$

`x`B$x<0$

`x`<0C$x>0$

`x`>0DFind the value of $y$

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

`x`when $y=3$`y`=3.

The function $y=\log x$`y`=`l``o``g``x` has been transformed into the function $y=5\log\left(x+4\right)-2$`y`=5`l``o``g`(`x`+4)−2.

Complete the following:

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

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

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