Natural Logs with the L Scale

Over forty years ago now, I had a girlfriend whose father was a carpenter. He was sort of a rustic, gruff type, really didn't care for society or the modern world very much. After becoming completely fed up with things, he bought a little parcel way up north, very close to Canada and proceeded to build a cabin in the woods for permanent homesteading. But he wanted to do it on his own terms.

Using just the minimum of hand tools, he started by constructing the other tools he would need to complete the job. Not a religious chap, still he was definitely working by an Amish ethic: no stinking 110V electricity need apply! As far as I know, he did it, too, and proceeded to live a rustic life in the wilds.

Well, sometimes that's the way I feel about slide rules. I take great delight in seeing what I can squeeze out of the humblest instrument. And that's what this installment is all about: how to find natural logs when all you have at the ready is the L scale, which just about every slide rule sports. The next installment shows you how to do it with the Ln scale, but that's almost too easy!

The basic idea here is to employ the old change-of-base formula. We'll go from common logs (on the L scale) to natural logs (base e) by one simple division. To be specific:

ln(x) = log(x) / log(e)

And log(e) is about 0.434, a conversion factor well worth memorizing. (This might be a good time to mention that you can use the methods of this section to find logs of any legal base; the natural base e just happens to be the one you'll need most frequently).

The change-of-base technique utilized here requires a number transfer. That is, you'll need to relocate a number from the L scale to the D scale. This always opens things up for imprecision, so be as tidy as you can. However, given that the alternative is using the Ln scale directly, and that Ln is so finely divided that it taxes the eyes no end, I think the difference in accuracy is negligible.

Be sure you've mastered the business of characteristics and mantissas treated in the previous three installments before proceeding. If you're all set, let's get going.

Exercise 1.1: Compute ln(3.5).

This'll be easy since 3.5 is one of the "basic numbers" I've mentioned before, i.e., it lies between 1.0 and 10.0 on the D scale, and yields a common log result (needed for the change-of-base) between 0 and 1.0 on the L scale.
  1. Set the cursor to 3.5 on the D scale.
  2. Read the number under the hairline on the L scale: 0.545. This is the common logarithm of 3.5.
  3. Move the cursor to 0.545 (5.45) on the D scale. This is the number-transfer alluded to above.
  4. Align 0.434 of the C scale with the hairline.
  5. Move the cursor to the left index.
  6. Read the result under the hairline on D: 0.125.
Here's how it appears on my favorite Pickett 1006-ES pocket slide rule. You can click the photos to enlarge them for details.

Cursor to D:3.5, read L:0.545
Cursor to D:0.545, slide C:0.434 to hairline, cursor to left index, result at D:0.125

This went without a hitch, because 3.5 was one of the basic numbers. For arguments greater than 10.0, you'll need to factor in the characteristic.

Exercise 1.2: Compute: ln(47.5).

You'll note the argument is one order of magnitude greater than that of the previous problem.
  1. Set the cursor to 47.5 (4.75) on the D scale.
  2. Read the number under under the hairline on the L scale: 0.676. This is the mantissa.
  3. Add the characteristic 1 (since 47.5 = 4.75 * 10^1), yielding 1.676.
  4. Set the cursor to 1.676 on the D scale.
  5. Align 0.434 of the C scale with the hairline.
  6. Move the cursor to the right index.
  7. Read the result under the hairline on the D scale: 3.86.
As it appears on my slide rule:

Cursor to D:4.75, read L:0.676, with characteristic is 1.676
Cursor to D:1.676, align C:0.434, cursor to right index, result at D:3.86

I'm assuming you had no problem placing the decimal point properly at the end. By estimation, 1.676/0.434 must be in the neighborhood of 2.0/0.4 = 5.

Really, that wasn't all that tedious of a calculation to carry out, and the result is actually good to 2 decimal places. We can do natural logs on even a Rietz type slide rule now!

But, what about arguments smaller than 1?

Exercise 1.3: Compute ln(0.8).

You'll recall that arguments less than 1 will make the common log, and hence the natural log after that division by 0.434, negative. Further keep in mind that the argument is now pinpointed on the DI scale (or CI scale with the rule closed), and that the log will be one order of magnitude less than for the basic numbers. See Common Logs with the L Scale if you need some review.

I'll give you a heads-up for this problem. In step 3 below, observe that since the common log is -0.01 (i.e., -1 from the slide rule's point of view) we don't have to divide by our usual scaling factor of 0.434, but can instead just find the reciprocal via CI. That saves a little time and slide movement.
  1. Set the cursor to 0.8 (8.0) on the CI scale.
  2. Read 0.1 on the L scale. Remembering the order of magnitude business for common logs of small arguments, this actually represents -0.01 in the long run.
  3. Rather than dividing 0.434 into -1, simply move the cursor to 0.434 on the C scale with the slide closed.
  4. The result is under the hairline on the CI scale: -0.23
Graphically, we have:

Cursor to CI:0.8, read L:-0.01
Cursor to C:0.434, result at CI:-0.23

The true result is actually closer to -0.22, but given that we approximated log(0.8) as -0.1 and moreover had to move the slide, we still did pretty well.

Okay, okay, that may have taken a little finagling, but still we did it! We found a natural logarithm using only what's available on a simple slide rule. Sort of like building a cabin in the woods with nothing more than a hammer and saw.

Next installment: Natural Logs with the Ln Scale