I'll be using the Ln scale exclusively, which appears on the Pickett N1006-T and Pickett N4-ES in my collection. I'll feature the former in all the pictures here.
Recall that the domain of e^x is all real numbers, while the range is all real numbers greater than 0. On a slide rule, the so-called "basic numbers" are 0.0 to 2.30, measured on the Ln scale, corresponding to their associated powers of e, 1.0 to 10.0, on the D or C scales. To say it more symbolically:
0 < x < 2.30 corresponds to 1.0 < e^x < 10.0
As in the last installment, it's worthwhile memorizing a basic relationship, in this case e^2.30 = 10.0. Of course, that 2.30 is rounded for slide rule purposes, the true value being irrational.
With that, let's try some problems.
Exercise 1.1: Find e^1.5.
The argument is one of the basic numbers (between 0.0 and 2.30) so we can proceed immediately.
- Set the cursor at 1.5 on the Ln scale.
- Read the result under the hairline on the D scale: 4.48.
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| Cursor to Ln:1.5, result at D:4.48 |
That was pretty straightforward, because 1.5 < 2.30. But what if we're outside of the basic numbers? Keep reading.
Exercise 2.1: Find e^3.5.
The argument is obviously larger than we want. However, using basic properties of exponents, we can split this up as:
e^3.5 = e^(2.30 + 1.20) = e^2.30 * e^1.20
In short, we strip off that power of 2.3, which is easily done by mental subtraction. And of course, the reason for doing so is because e^2.30 = 10, a very convenient number indeed. In short, our result will be given by:
10 * e^1.20
and, of course 1.20 is one of the basic numbers, so we can proceed as in Exercise 1.1.
- Set the cursor at 1.20 on the Ln scale.
- Read the number under the hairline on the D scale: 3.31.
- Multiply by 10 to yield: 33.1.
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| Set cursor at Ln:1.20, read D:3.31, multiply by 10:33.1 |
And that's an accurate result to one decimal place.
Detaching more groups of e^2.30 is certainly permissible, as the next example shows.
Exercise 2.2: Find e^10.
Realizing that 10 is not in the range of our basic numbers, we split off 2.30. But that leaves too much left over, so remove another 2.30. Still no good, and this is getting tiresome. So instead, consider some ordinary long division with remainder as taught in elementary schools:
10 / 2.30 = 4 * 2.30 + 0.8
Hence,
e^10 = e^(4 * 2.30 + 0.8) = (e^2.30)^4 * e^0.8
While that may look intricate or messy, it really isn't. We've simply determined that there are four complete groups of e^2.30, each equal to 10, with an e^0.8 left over. And, of course, 0.8 is one of our basic numbers. Thus, the desired result will be:
e^0.8 * 10^4
Let's do it.
- Set the cursor at 0.8 on the Ln scale.
- Read the number under the hairline on the D scale: 2.23.
- Multiply by 10^4 (10000) to yield: 22,300.
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| Cursor to Ln:0.8, read D:2.23, multiply by 10^4: 22300 |
Now if you check this against a calculator, you'll see a result of about 22026, so our value seems a bit high. What gives? Well, it's that rounded 2.30 we used. Appearing only once, as in Exercise 2.1, it performed well enough. But employ it 4 times sequentially and the error accumulates. In particular, the remainder in the exponential manipulation above is better stated as 0.79, not 0.80. Doing so, results in just a hair over 22,000, a much closer value.
However, given that e^x grows so rapidly, it's hardly worth the effort for most engineering applications to laboriously carry out a long division with 2.302585... instead of 2.30 when reducing the exponent. Our 22300 is within 1%, which surely is nothing to grouse about when performing analog computation. Anyway, the Ln scale is quite finely divided and fairly hard to read (at least with my aging eyes), so I think it's diminishing returns to worry about the 2.30 factor.
We really ought to try a negative argument.
Exercise 3.1: Find e^(-0.66).
Having read the previous entries here, you won't be surprised to learn we'll be using the DI or CI scales, instead of D. I'll remind you, if you don't have DI on your slide rule, you can always use CI, just by closing the rule (aligning the indices).
Since we're talking reciprocals, the basic numbers drop by an order of magnitude (and we just ignore the minus sign when using the Ln scale). To be specific:
-2.30 < x < 0 corresponds to 0.1 < e^x < 1.0
We saw something similar in Powers of 10. It's simply that the decimal point has moved one place over. Let's get going on the problem.
- Set the cursor at 0.66 (-0.66) on the Ln scale.
- Read the result under the hairline on the DI (or CI) scale: 0.517.
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| Cursor to Ln:0.66, result at CI:0.517 |
It'll be no great surprise that we can strip off groups of 2.30 with negative exponents, too.
Exercise 3.2: Find e^(-7.3).
Rewrite this as: e^(-7.3) = (e^7.3)^(-1) = (e^2.30 * 3)^(-1) * e^(-0.4).
We see that the first factor is 10^(-3), and the second factor has its argument in the basic number range for negatives, one order of magnitude less; keep that in mind.
- Set the cursor at 0.4 on the Ln scale.
- Read the number under the hairline on the CI scale: 0.67 (remember: one order of magnitude less, since the argument was negative).
- Multiply by 10^(-3): 0.67 * 10^(-3) or more properly 6.7 * 10^(-4).
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| Cursor to Ln:0.4, read CI0.67, scale:6.7*10^(-4) |
A digital calculator shows a result of 6.8 * 10^-4). Had we wielded a more precise value than 2.30 when stripping things down we would have looked for 0.39 on the Ln scale, not 0.4. Nonetheless, our humble analog result is still less than 1% off.
I'm going to be the first to say, finding powers of e with just the Ln scale at one's disposal is far from slick. That business of stripping off groups of 2.30 is a trifle laborious. The log-log scales make much shorter (and more accurate) work of this type of problem.
But, and this is the important point, we can still do it! My little N1006-T slide rule which fits in a breast pocket is able to compute all manner of natural logs and powers of e, even if I do have to furl my brow to think things through.
However, as I mentioned above, not all slide rules have even an Ln scale. But not to worry, in the next installment I'll show you how to handle these computations with the natural base even if all you have is an L scale.
Next installment: Natural Logs with the LL Scales