But viewing this instead as an antilogarithm all of a sudden makes its utility clear. In the manner of Henry Briggs (the first advocate of common logs), we'll find there are a number of interesting applications in which we first convert expressions to logarithms, do some simple arithmetic, then convert that sub-tally back to a desired result via the antilogarithm.
Last time we agreed to call the span of numbers on the C or D scale the basic range, i.e., 1.0 through 10.0, with the corresponding mantissas on the L scale covering 0.0 through 1.0.
Now with powers of 10, we'll reverse gears but still call it the basic range. In short, on a slide rule the arguments for powers of 10 will lie between 0.0 and 1.0, with a result in the range 1.0 to 10.0. Dealing with arguments outside of that span is easy, just by remembering a simple property of exponents. Let's get started.
Arguments Between 0 and 1
In all of the problems to follow, you'll be starting with the L scale, then working backwards to the D scale. Depending on the arrangement of your slide rule (is L on the slide or stator?) you might be using C instead. And where reciprocals come into play, you might need CI instead of DI.
Exercise 1.1: Find 10^0.3.
- Set the cursor at 0.3 on the L scale.
- Read the result under the hairline on D: 1.995.
 |
| Cursor to L:0.3, result at D:1.995 |
Exercise 1.2: Find 10^(0.57).
- Set the cursor at 0.57 on the L scale.
- Read the result under the hairline on D: 3.72.
 |
| Cursor to L:0.57, result at D:3.72 |
If you stop to think about it, we now know how to find any simple root of 10, i.e., square root, cube root, fourth root, etc.
Arguments Greater Than 1
This is really easy, recalling how addition of exponents works. For example, given 10^4.7, split this up into (10^4) * (10^0.7). You'll notice that the second factor now has an argument falling in the basic range which can be worked directly using the method just seen. Just keep the 10^4 as your order of magnitude.
And of course, once you know the reason, there's no need to actually think of the exponent rule used: simply split the number 4.7 into a 4 and a 0.7 and carry on.
Exercise 2.1: Find 10^4.7.
- Set the cursor at 0.7 on the L scale.
- Read the number under the hairline on the D scale: 5.
- Form the result: 5 * 10^4.
 |
| Cursor to L:0.7, read number D:5, form result 5*10^4 |
Exercise 2.2: Find 10^21.5.
- Set the cursor at 0.5 on the L scale.
- Read the number under the hairline on the D scale: 3.16.
- Form the result: 3.16 * 10^21
 |
| Cursor to L:0.5, read number D:3.16, form result 3.16*10^21 |
You'll note in these problems, the result is automatically shaped into scientific notation form. Nice!
Arguments Between -1 and 0
Recalling that negative arguments imply a reciprocal is on the menu, you won't be startled to learn problems like this use the DI (or CI) scale instead of D (or C).
Exercise 3.1: Find 10^(-0.75).
- Set the cursor at 0.75 on the L scale.
- Read the result under the hairline on CI: 0.178.
 |
| Cursor to L:0.75, result at CI:0.178 |
Now, you might be nonplussed that we ended up with 0.178 and not 1.78. After all, isn't the latter in that basic range alluded to earlier? It would be but for the fact the exponent was negative, implying a reciprocal was a-coming. So, if the basic range is 1.0 to 10.0, then the multiplicative inverses of that range (found on CI) span 1.0 down to 0.1.
Exercise 3.2: Find 10^(-0.2).
- Set the cursor at 0.2 on L.
- Read the result under the hairline on CI: 0.631
 |
| Cursor to L:0.2, result at CI:0.631 |
Once again, we arrive at a number between 0.1 and 1.0.
Arguments Less Than -1
A moments reflection ought to convince you that one can combine the techniques from above to arrive at powers of 10 when the argument is less than -1. For example: 10^(-12.5) = 10^(-12) * 10^(-.5). Just press ahead with what you've already learned and you'll arrive at 3.16 * 10^(-13).
Pop Quiz: Do you understand why that exponent is -13? Think about the standard form for scientific notation.
Tiny Arguments
If the argument shrinks toward zero, then clearly the power of 10 approaches 1. In fact, it'll get so close on a slide rule that you probably ought to give up once the argument drops below 0.01. Nonetheless, just for the heck of it, let's try one that's small but not so small that we can't read it under the hairline.
Exercise 4.1: Find 10^0.038.
Now this is tricky, but do-able. We need to get things into the basic range first. So write the expression as 10^0.038 = 10^(1-0.962).
Okay, finding the complement of 0.038 as 0.962 isn't exactly automatic, but it's still a reasonable mental computation. Anyway, we can now carry on with the methods from above.
- Set the cursor at 0.962 on L.
- Read the number under the hairline on CI: 0.109.
- Adjust result to 1.09 since the characteristic was 1.
 |
| Cursor to L:0.962, result at CI:1.09 |
To repeat, we were working with an exponent of 1- 0.962 at the outset; that 1 represents the characteristic, one complete power of 10. So, at the end, we needed to factor it back in.
The domain of 10^x is all real numbers, and in the paragraphs above we have accounted for all possibilities, realizing that it always comes back to what's happening in the basic range.
Next installment: Arbitrary Powers with L Only