Graphing Logarithmic Functions: Introduction


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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 square-root graph:

*y = sqrt(x)*	*y = log₂(x)*

But, while the square root starts at the point (0, 0) and then goes off to the right, the log passes through (1, 0), going off to the right but also sliding down the positive side of the y-axis. Remembering that logs are the inverses of exponentials, this shape makes perfect sense, since the graph of the log, being the inverse of the exponential, would just be the flip of the graph of the exponential:

y = 2^x	*y = log₂(x)*

comparison of the two graphs, showing the inversion line in red

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 y-values, and plot the points. But how do you graph logs? There are two options. Here is the first: Copyright © Elizabeth Stapel 2006-2008 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, so I won't even bother trying to find points for x = –3 or x = 0, for instance. So I'll start with x = 1, and work from there:

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, I won't bother with 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 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.

This gives me the point (1, 0) and some points to the right, but what do I do for x-values between 0 and 1? For this interval, I need to think in terms of negative powers and reciprocals. Just as the left-hand side of the exponential function had few graphable points (the rest of them being too close to the x-axis), so also the bottom of the log function has few graphable points, the rest of them being too close to the y-axis. But I can find a few:

Since 2^–1 = ¹/₂ = 0.5, then log₂(0.5) = –1, and (0.5, –1) is on the graph. Since 2^–2 = ¹/₄ = 0.25, then log₂(0.25) = –2, and (0.25, –2) is on the graph. Since 2^–3 = ¹/₈ = 0.125, then log₂(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 which I view as being too small to bother with, so I'll quit with the values I've got.