Graphing Logarithmic Functions: Intro (page 1 of 3) By nature of the logarithm,
most log graphs tend to have the same shape, looking similar to a squareroot
graph:
The graph of the square root starts at the point (0, 0) and then goes off to the right. On the other hand, the graph of the log passes through (1, 0), going off to the right but also sliding down the positive side of the yaxis. 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:
It is fairly simple to graph exponentials. For instance, to graph y = 2^{x}, you would just plug in some values for x, compute the corresponding yvalues, and plot the points. But how do you graph logs? There are two options. Here is the first: Copyright © Elizabeth Stapel 20022011 All Rights Reserved
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 yvalues for, say, x = –3 or x = 0. Instead, I'll start with x = 1, and work from there, using the definition of the log.
The above gives me the point (1, 0) and some points to the right, but what do I do for xvalues between 0 and 1? For this interval, I need to think in terms of negative powers and reciprocals. Just as the lefthand "half" of the exponential function had few graphable points (the rest of them being too close to the xaxis), so also the bottom "half" of the log function has few graphable points, the rest of them being too close to the yaxis. But I can find a few: Since 2^{–1} = ^{1}/_{2} = 0.5, then log_{2}(0.5) = –1, and (0.5, –1) is on the graph. Since 2^{–2} = ^{1}/_{4} = 0.25, then log_{2}(0.25) = –2, and (0.25, –2) is on the graph. Since 2^{–3} = ^{1}/_{8} = 0.125, then log_{2}(0.125) = –3, and (0.125, –3) is on the graph. The next power of 2 (as x moves in this direction) is ^{1}/_{16} = 2^{–4}, but the xvalue for the point (0.0625, –4) seems too small to bother with, so I'll quit with the points I've already found.
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