You'll be glad. (And of course, bear in mind what I wrote in Some Terminology about the meaning of "equals" in analog computation).
With that out of the way, let's get going and see how to compute natural logarithms, that is, logarithms with base e. In this installment, we'll be using the Ln scale only (not the log-log scales found on fancier and more costly slide rules). Now Ln is not found all that commonly. Among the slide rules in my collection, only two sport it: the Pickett N1006-T, and the Pickett N4-ES. That it appears on the former is particularly attractive. This pocket unit is virtually identical to the 1006-ES I've been showing throughout here, with two important additions. First, it also includes the Ln scale. Then, all of the inverse scales are printed in red. So, this handy and pleasing rule will likely become one of my real favorites.
The place to begin with the natural logarithm is by considering its domain and range.
The domain of ln(x) is all positive real numbers. The range is all real numbers. In a manner quite similar to how we computed Common Logs, we'll find an argument on the C, D, CI or DI scale, then read the natural log from the Ln scale, adjusting the characteristic as required.
The arguments, running from 1.0 to 10.0, correspond to values of 0.0 through 2.30. We'll call these the "basic numbers," in much the same way we did for the common logarithms. These are the numbers you'll find on the opposing scales, requiring no mental adjustment for order of magnitude. With that in mind, we can work our first problem.
Exercise 1.1: Find ln(2.5).
With the Pickett N1006-T slide rule I'm using, the argument is pegged on the D scale, and the associated natural log on the Ln scale. Both appear on the stator (base or stock).
We see that 2.5 is one of the basic numbers, so we can jump right in.
- Set the cursor to 2.5 on the D scale.
- Read the result under the hairline on the Ln scale: 0.916.
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| Cursor to D:2.5, result at Ln:0.916 |
But what about arguments greater than 10.0, or for that matter less than 1.0? I'm glad you asked. Inspect the following figure.
In Exercise 1.1, the argument lay in the basic number span (1.0 to 10.0) which was set on the D scale. The result lay in interval 0.0 to 2.30 on the Ln scale.
Now suppose, we're searching for ln(37.2). That falls in the next interval to the right, between 10.0 and and 100.0. But consider this:
ln(37.2) = ln(10 * 3.72) = ln(10) + ln(3.72)
Of course, you already know ln(10), because I'm sure you took my advice above and memorized it, right? And 3.72 is one of the basic numbers we already know how to deal with. After a trivial sum, you'll have the desired result! Let's try it.
Exercise 1.2: Find ln(37.2).
We'll split the argument as sketched out above into a characteristic and mantissa, then press ahead.
- Set the cursor at 3.72 on the D scale.
- Read the number under the hairline on the Ln scale: 1.32.
- Add this to 2.30 for the result: 3.62.
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| Cursor to D:3.72, read Ln:1.32, result is 1.32+2.30=3.62 |
Exercise 1.3: Find ln(455).
Referring to the figure above, it should be clear that we can extend this process indefinitely to the right. Every time the argument jumps to the next order of magnitude, simply add on another 2.30. Truthfully that number is closer to 2.30256, so doubling it yields 4.61, rounded. If you'd like it spelled out for this example:
ln(455) = ln(10 * 10 * 4.55) = ln(10) + ln(10) + ln(4.55)
You'll note that 4.55 is, once more, a basic number.
- Set the cursor to 4.55 on the D scale.
- Read the number under the hairline on the Ln scale: 1.51.
- Add 2.30 twice (more precisely, 4.61 once) to arrive at: 6.12.
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| Cursor to D:4.55, read Ln:1.51, result is 1.51+4.61=6.12 |
But let's turn around and march in the opposite direction. Referring to the figure above once more, we see that for arguments below 1.0 (but greater than 0, obviously), we'll get a negative result, as expected. But think about this for a moment: arguments between 0 and 1.0 are proper fractions, and proper fractions are easily formed by the DI or CI scales.
Hence, instead of locating the argument on D, find it on DI instead, read the result off of Ln as usual, finally prefixing a minus sign. Nothing tricky at all!
One note: for a slide rule not featuring DI, simply close up the rule (aligning C with D) and use CI instead. That's what I'll be doing with the Pickett N1006-T in the pictures here, simply because CI is on the same side as Ln (DI is on the flip-side). No sense wasting energy on a duplex-flip...
Exercise 2.1: Find ln(0.25).
Here we go with a fractional argument, but larger than 0.1.
- Set the cursor at 0.25 (2.5) on the CI scale.
- Read the number under the hairline on the Ln scale: 1.386.
- Prefix a minus sign: -1.386.
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| Cursor to CI:2.5, read Ln:1.386, negate:-1.386 |
And there's no reason why we can extend this protocol further to the left.
Exercise 2.2: Find ln(0.017).
Here we have an argument one order of magnitude smaller than before. So, we'll simply subtract our magic number, 2.30, at the end.
- Set the cursor at 0.017 (1.7) on the CI scale.
- Read the number under the hairline on the Ln scale: 1.77.
- Negate this: -1.77.
- Subtract 2.30 to arrive at -4.07.
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| Cursor to CI:1.7, read Ln:1.77, negate, subtract 2.30:-4.07 |
Now nobody is particularly fond of subtraction, but note the following. Rather than negating that 1.77 and then subtracting 2.30, why not simply add 1.77 and 2.30, then negate. Same difference, and of course it's an easy mental sum. Pretty clearly, this attack applies to any order of magnitude smaller you desire; just keep adding another 2.30 each time you drop a notch, then prefix a minus sign at the very end.
Well, I don't know about you, but I think this is all pretty cool! This little baby slide rule is able to compute any natural log, no matter the size of the argument. I still marvel that something so small, with no batteries and but two moving parts can compute products, quotients, inverses, square and cube roots, all trig functions, common logs and now natural logs...it ranks as one of mankind's most amazing inventions.
Next installment: Powers of e with the Ln Scale