To review: below are some different variations on the same ...



Graphing Logarithmic Functions

Remembering that logs are the inverses of exponentials, this shape for the log graph makes perfect sense: the graph of the log, being the inverse of the exponential, would just be the "flip" of the graph of the exponential:

|y = 2x |y = log2(x) |

|[pic] |[pic] |

|comparison of the two graphs,  |

|showing the inversion line in red |

|[pic] |

It is fairly simple to graph exponentials. For instance, to graph y = 2x, you would just plug in some values for x, compute the corresponding y-values, and plot the points. But how do you graph logs? There are two options. Here is the first:   Copyright © Elizabeth -2011 All Rights Reserved

• Graph y = log2(x).

In order to graph this "by hand", I need first to remember that logs are not defined for negative x or for x = 0. Because of this restriction on the domain (the input values) of the log, I won't even bother trying to find y-values for, say, x = –3 or x = 0. Instead, I'll start with x = 1, and work from there, using the definition of the log.

• Since 20 = 1, then log2(1) = 0, and (1, 0) is on the graph.

• Since 21 = 2, then log2(2) = 1, and (2, 1) is on the graph.

• Since 3 is not a power of 2, then log2(3) will be some messy value. So I won't bother with graphing x = 3.

• Since 22 = 4, then log2(4) = 2, and (4, 2) is on the graph.

• Since 5, 6, and 7 aren't powers of 2 either, I'll skip them and move up to x = 8.

• Since 23 = 8, then log2(8) = 3, so (8, 3) is on the graph.

• The next power of 2 is 16: since 24 = 16, then log2(16) = 4, and (16, 4) is on the graph.

• The next power of 2, x = 32, is too big for my taste; I don't feel like drawing my graph that wide, so I'll quit at x = 16.

Graph the point to create the graph of y = log2(x)

[pic]

• Graph y = log3(x) + 2.

This is the basic log graph, but it's been shifted upward by two units. To find plot points for this graph, I will plug in useful values of x (being powers of 3, because of the base of the log) and then I'll simplify for the corresponding values of y.

30 = 1, so log3(1) = 0, and log3(1) + 2 = 2 

31 = 3, so log3(3) = 1, and log3(3) + 2 = 3 

32 = 9, so log3(9) = 2, and log3(9) + 2 = 4 

33 = 27, so log3(27) = 3, and log3(27) + 2 = 5

|   |   |[pic] |

|The graph of y = log3(x) + 2looks like | | |

|this: | | |

• Graph y = log2(x + 3).

This graph will be similar to the graph oflog2(x), but it will be shifted sideways.

Since the "+ 3" is inside the log's argument, the graph's shift cannot be up or down. This means that the shift has to be to the left or to the right. But which way? You can keep track of the direction of the shift by looking at the basic point(1, 0) ("basic" because it's neat and easy to remember). The log will be 0 when the argument,x + 3, is equal to 1. When is x + 3 equal to 1? When x = –2. Then the basic log-graph point of (1, 0) will be shifted over to 

(–2, 0) on this graph; that is, the graph is shifted three units to the left. If you are not comfortable with this concept or these manipulations, please review how to work with translations of functions.

Since a log cannot have an argument of zero or less, then I must have x + 3 > 0, this tells me that, for this graph, x must always be greater than –3.

The graph of the basic log function y = log2(x) crawled up the positive side of the y-axis to reach the x-axis, with the line never going to the left of the limitation that x must be greater than zero. To remind myself of the similar limitation of this log (where x must always be greater than–3), I will insert a dashed line at x = –3:

[pic]

A line like this, which marks off territory where the graph shouldn't go, is called a "vertical asymptote", or simply an "asymptote". I don't have to add this to the graph, but it can be very helpful, and might convince the grader that I know what I'm doing.

After I dash in the asymptote, I plot some points:

20 = 1, so log2(1) = 0; x + 3 = 1 for x = –2:  (–2, 0) 

21 = 2, so log2(2) = 1; x + 3 = 2 for x = –1:  (–1, 1) 

22 = 4, so log2(4) = 2; x + 3 = 4 for x = 1:  (1, 2) 

23 = 8, so log2(8) = 3; x + 3 = 8 for x = 5:  (5, 3)

|   |  |[pic] |

|Plotting the points I've calculated, I get: | | |

|   |  |[pic] |

|...and connecting the dots gives me the following | | |

|graph: | | |

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3.2.- 3.3. Graphing Logarithmic Functions Practice.

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