Arbitrary Powers with L Only

Being able to compute a number to an arbitrary power (or take an arbitrary root) is an important skill needed by any scientific practitioner. Slide rules can do it!

Now I'm not going to kid you; raising an argument to an arbitrary number, or taking an arbitrary root of that argument is a task best left to the log-log scales...if you have them. More powerful slide rules sporting those scales make short work of such problems.

But one of the reasons I've written up this blog of techniques is to see just how much can be wrung out of a simple slide rule, like the Pickett 1006-ES I favor. With that in mind, this installment covers a method for computing powers and roots just wielding the L scale along with the usual culprits. It can be done, but requires two number-transfers, which always open up the possibility of accumulated imprecision.

I like to think of it this way. Suppose I were marooned on a desert island (which has always struck me as a curious term) and needed to find 3.5 raised to the 2.7, or the fourth root of 5. I would obviously prefer to have a Post Versalog at hand to crank things out on the log-log scales. But what if when I was stranded, only my trusty Pickett pocket slide rule survived. I could still press ahead, even if needing a couple extra steps.

So simply in the interest of squeezing as much as we can out of a basic slide rule, let's see how to conquer these more complex problems with L, a scale which appears on just about any unit.

What It's Based Upon


You'll recall from the previous two installments, we agreed to use the term basic range to mean the following. If we think of the C (or D) scale as indicating numbers between 1.0 and 10.0, then the corresponding common logarithms on the L scale span 0.0 to 1.0. (If you're a stickler for details, note that the right endpoint of either sets you back to the start again, but with a characteristic of 1 now).

For numbers outside of the basic range, either smaller or larger, we merely bring the ideas of characteristic and order of magnitude into play. Again, please refer to the previous two installments for details.

So, when tackling these new conundrums involving arbitrary exponents and roots, we'll continually reduce the numbers to lie within the basic range, simply maintaining the characteristic mentally. It's actually pretty straightforward once you've worked a couple practice problems.

The basic approach is simplicity itself, thanks to John Napier. In particular: log(b^n) = n * log(b). Crank out the right side (both the multiplication and the log are easy to do on a slide rule), then find the antilogarithm (again an easy task) and you're done.

Finding a nasty root? Then, log[b^(1/n)] = log(b)/n, take the antilogarithm and you've got it.

One last thing. Some rules put L on the slide, some on the stator. Thus you may be comparing it against C in the first case, or D in the second. A similar situation applies to CI and DI. So, in what follows, you may need to adjust the instructions accordingly depending on what style slide rule you're using.

Let's try some problems.

Argument Raised to an Arbitrary Power


Exercise 1.1: Find 3.5^2.7.

We know that the first step is to find the common log of 3.5. I'll ask you to note that this argument lies in the basic range, so we can proceed at once. However, I'll expand the instructions a bit more than in previous installments for extra detail.
  1. Set the cursor at 3.5 on the D scale. This is the base.
  2. Read the number under the hairline on the L scale: 0.544. This is the logarithm of the base.
  3. Move the cursor to 0.544 on the D scale.
  4. Align 2.7 (the exponent) of the CI scale with the hairline.
  5. Move the cursor to the left index and read the product under the hairline on D: 1.47. This is the product log(3.5) * 2.7.
  6. The 1 in 1.47 is the characteristic (remember it), while 0.47 is the logarithm lying in the basic range.
  7. So, move the cursor to 0.47 on the L scale.
  8. Read the antilogarithm under the hairline on the D scale: 2.94.
  9. Finally, factor in that characteristic (order of magnitude) that you remembered to arrive at the final result: 29.4
That all sounds a lot worse than it really is as the following pictures indicate:

Cursor to D:3.5, read log L:0.544
Cursor to D:0.544, CI:2.7 to hairline, read left index D:1.47
Cursor to L:0.47, read D:2.94, interpret as 29.4

I hope it's obvious that in step 4 we were using the old "divide by a reciprocal" trick with CI to carry out the multiplication.

Run through the steps for practice a couple times, noting how that 1 in step 6 got stripped away to that we could continue on with numbers in the basic range.

In the next problem you'll need to consider the basic range at the start as well. It really isn't all that big of a deal after just a modicum of practice.

Exercise 1.2: Find 15^5.2.
  1. Set the cursor at 1.5 (which with a characteristic of 1 represents 15) on the D scale.
  2. Read the logarithm of 1.5 under the hairline on the L scale: 0.176.
  3. Append the characteristic from step 1, yielding the complete logarithm of 1.176.
  4. Set the cursor to 1.176 on the D scale.
  5. Align 5.2 of the CI scale with the hairline.
  6. Move the cursor to the right index.
  7. Read the product under the hairline on the D scale: 6.12.
  8. Thus, the characteristic of this log is 6; remember it.
  9. Move the cursor to 0.12 on the L scale.
  10. Read the antilogarithm under the hairline on the D scale: 1.31.
  11. Factor in that characteristic of 6 (i.e., scale by 10^6): 1,310,000.
In pictures, you'll see:

Cursor to D:1.5, read log L:0.176, form complete log 1.176
Cursor to D:1.176, align CI:5.2 with hairline, read right index D:6.12
Cursor to L:0.12, read D:1.31, interpret as 1310000

I hope I haven't freaked you out with this detailed description of the steps. True, you'll be doing a couple number-transfers, but I can promise you after you've run through it all a few times, it really does become second nature if you'll only remember Napier's power rule.

Arbitrary Roots


We need to try a root problem. Everything runs pretty much the same way, except we'll be dividing that common log early on, not multiplying into it.

Exercise 1.3: Find the fourth root of 5.
  1. Set the cursor at 5 on the D scale.
  2. Read the logarithm under the hairline on the L scale: 0.7.
  3. Move the cursor to 0.7 on the D scale.
  4. Align 4 on the C scale with the hairline.
  5. Move the cursor to the left index.
  6. Read the quotient under the hairline on the D scale: 0.175.
  7. Move the cursor to 0.175 on the L scale.
  8. Read the result under the hairline on the D scale: 1.5.
You'll note in this problem that everything stayed within the basic range. Pictorially, we have:

Cursor to D:5, read logarithm L:0.7
Cursor to D:0.7, align C:4 with hairline, read left index D:0.175
Cursor to L:0.175, result (antilogarithm) at D:1.5

(While writing this up, it occurred to me that for this particular problem it would be faster and probably more accurate to simply take the square root of 5 twice with the A scale. So, keep your eyes open!)

I'll confess, there's a fair amount of slip-sliding going on with these problems. But don't you think it's cool that one can even attempt arbitrary roots and powers with just the simplest of slide rules? Give me C, D, CI, DI and L and I'm ready to conquer the world...

Next installment: Natural Logs with the L Scale