Logs of Arbitrary Base by LL Scales

Man, those LL scales; they really are something! In the last two installments, we saw how they could be used to find natural logs and powers of e. Yeah, pretty cool, but how 'bout using them to determine logarithms of any legal base! That'll be the thrust of this section. Very powerful, indeed...

Now, a bit about the notation. I'm not really sure how to do subscripts decently on a Web site like this, so I'll use the following schema:

log(b, x)

to mean "the logarithm base b of x." So, for example, log(4, 16) = 2 should be read as "the logarithm, base 4, of 16 is equal to 2." This, of course, implies that 4^2 = 16, as usual.

To get started, let me describe how to find such a logarithm, stating the steps as baldly as possible. Then we'll take a look at the machinery that makes it work.

To find log(4, 16):
  1. Set the cursor at 4 on the LL3 scale.
  2. Align the left index with the hairline.
  3. Move the cursor to 16 on the LL3 scale.
  4. Read the result, 2, under the hairline on the C scale.
That was mighty efficient, wasn't it! But how on earth does it work? Examine the following figure:

Click to enlarge

Think of this as a multiplication problem, using the canonical approach we learned in the very first tutorial here:

The left index (C:1) is aligned with 4 on the LL3 scale, but notice that this corresponds to about 1.386 on the D scale. In other words, we've just set things up to multiply by ln(4) ≈ 1.386.

Next, suppose we wanted to multiply that number by 2. Observe that the cursor has been moved rightward until it aligns with 2 on the C scale. 2, of course signifies log(4, 16), which strictly speaking we're not supposed to know yet. But stay with me...

Under the hairline, on the D scale is a result of 2.773. But this length maps to ln(16) on the LL3 scale.

Putting these segments all together, still thinking of the total arrangement as a multiplication problem:

ln(4) * log(4, 16) = ln(16)

Rearranging the relationship, we arrive at:

log(4, 16) = ln(16) / ln(4)

Aha! The old change-of-base gambit!

The method sketched out above is not only legitimate, but it works slick as hell. Let's see it in action.

Exercise 1.1: Find log(8, 37).

Both numbers here (the base 8 and the argument 37) appear on the LL3 scale.
  1. Set the cursor to 8 on the LL3 scale.
  2. Align the left index with the hairline.
  3. Move the cursor to 37 on the LL3 scale.
  4. Read the result under the hairline on the C scale: 1.74.
Remember that many slide rules, like the K+E 4181-1 I'm using here, label the right side of the scales with reminders. In this case, we see that the result should lie between 1.0 and 10.0, which in any event jibes with a quick estimation. Here's how it appears (again, click the photo if you want to see it full-sized):


Cursor to LL3:8, left index to hairline, cursor to LL3:37, result at C:1.74

Of course, we can use the other LL scales for smaller arguments.

Exercise 1.2: Find log(5, 1.5).
  1. Set the cursor to 5 on the LL3 scale.
  2. Align the left index with the hairline.
  3. Move the cursor to 1.5 on the LL2 scale.
  4. Read the result under the hairline on the C scale: 0.252.
The label on the rule reminds us that the result should lie between 0.1 and 1.0. In passing, I'll ask you to note the base is on the LL3 scale, the argument on the LL2 scale (one notch lower), with the result one order of magnitude smaller than that arrived at in Exercise 1.1. I'll explain in more detail later, but for now, whenever you skip from one LL scale to an adjoining one, the magnitude changes accordingly.

Graphically, we have:

Cursor to LL3:5, align left index, cursor to LL2:1.5, result at C:0.252

Let's try one with an even smaller argument. On my K+E here, I'll need to do a duplex-flip, since LL1 is on the reverse side of C. (Most slide rules tend to put two of the LL scales on one side, with the third opposite that).

Exercise 1.3: Find log(20, 1.07).
  1. Set the cursor to 20 on the LL3 scale. This is the base, of course.
  2. Align the left index with the hairline.
  3. Move the cursor to 1.07 on the LL1 scale.
  4. Read the result under the hairline on the C scale: 0.0266.
An order of magnitude smaller than the previous problem, but thanks to the helpful labels on the K+E 4181-1 slide rule, we knew that was coming. As mentioned, a couple duplex-flips are required (C and LL1 are on opposing sides), so it'll take three pictures to show it all:


Cursor to LL3:20, left index to hairline, duplex-flip
Cursor to LL1:1.07, duplex-flip
Result at C:0.0266

We should really try a couple problems in which the base is less than e, and hence needs to be located on LL2 or LL1.

Exercise 2.1: Find log(1.5, 48).
  1. Set the cursor to 1.5 on the LL2 scale.
  2. Align the right index with the hairline.
  3. Move the cursor to the 48 on the LL3 scale.
  4. Read the result under the hairline on the C scale: 9.55.
In summary:

Cursor to LL2:1.5, align right index, cursor to LL3:48, result at C:9.55

Did you see the right index coming to keep from going off-scale? This requires a mental adjustment, but it's simple enough to accommodate. With the base being located on the LL2 scale, we expect to interpret the C scale as 0.1 to 1.0. However, by using the right-index, we are instead thinking of the result being off to the imaginary right of the slide-rule, i.e., the next order of magnitude up. So, it's 9.55 and not 0.955.

This might be a good time to mention that the LL scales, though tripartite, really should be thought as connecting to each other, yielding one triple-length scale: LL1 + LL2 + LL3.

Exercise 2.2: Find log(1.2, 1.7).

In this problem, both b and x lie upon the LL2 scale. That'll be important to remember in just a moment.
  1. Set the cursor to 1.2 on the LL2 scale.
  2. Align the left index with the hairline.
  3. Move the cursor to 1.7, also on the LL2 scale.
  4. Read the result under the hairline on the C scale: 2.91.
In pictures,

Cursor to LL2:1.2, align left index, cursor to LL2:1.7, result at C:2.91

You may have been expecting 0.291, since we were using the LL2 scale. I mean, that's how we mentally adjusted things in Exercise 1.2. Here's the deal. It's only when "crossing boundaries" that the order of magnitude changes. In Exercise 1.2, we had the base, b, on the LL3 scale, but the argument, x, on the LL2 scale.

The general rule is: C maps to the range 1.0 to 10.0 when both b and x are on the same LL scale. But if x is located on a scale lower than that on which b lies, then the order drops.

Read over the following relationships and think about them for a moment. There's actually nothing to memorize since the pattern is fairly obvious.

LL3:b, LL3:x ---> 1.0 to 10.0
LL2:b, LL2:x ---> 1.0 to 10.0
LL1:b, LL1:x ---> 1.0 to 10.0

LL3:b, LL2:x ---> 0.1 to 1.0
LL3:b, LL1:x ---> 0.01 to 0.1

LL2:b, LL3:x ---> 0.1 to 1.0

We just seen what happens when we "drop" from one LL scale to the next below. And, of course, you can reverse gears as well and move upward.

To frame this in words, C is expected to give values of logarithms between 1.0 and 10.0. But, when moving from one LL scale for b, to the next higher adjacent one for x (say, going from LL1 to LL2) the order of magnitude is bumped up a notch, 10.0 to 100.0. Similarly, going from LL1 to LL3 puts you in the 100.0 to 1000.0 range. And clearly when moving downward (LL3 to LL2 to LL1), you'll decrement the order of magnitude, as we just saw.

Naturally, you'll also need to take into account whether you used the right index, as in Exercise 2.1. That has the effect of shifting things one way or another as well.

That may all sound messy, but trust me: sit quietly, work a few examples and look for the patterns; it'll quickly make eminent sense, I promise!