Anyway, we can now carry on with the methods from above. Okay, finding the complement of 0.038 as 0.962 isn't exactly automatic, but it's still a reasonable mental computation. We need to get things into the basic range first. 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. 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. If the argument shrinks toward zero, then clearly the power of 10 approaches 1. Pop Quiz: Do you understand why that exponent is -13? Think about the standard form for scientific notation. Just press ahead with what you've already learned and you'll arrive at 3.16 * 10^(-13). Once again, we arrive at a number between 0.1 and 1.0.Ī 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. Read the result under the hairline on CI: 0.631.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. 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. Now, you might be nonplussed that we ended up with 0.178 and not 1.78. In the photo here, I'm using CI since it lies on the same side as L, thus avoiding a duplex-flip: Read the result under the hairline on CI: 0.178.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). You'll note in these problems, the result is automatically shaped into scientific notation form.
#Power of ten rule how to#
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.
Dealing with arguments outside of that span is easy, just by remembering a simple property of exponents. 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. Now with powers of 10, we'll reverse gears but still call it the basic range. 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. 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. Now finding a power of 10 probably sounds exceedingly dull I mean how often do you run into such a problem in scientific calculations?īut viewing this instead as an antilogarithm all of a sudden makes its utility clear. Obviously, one can turn this around to compute a power of 10. In the previous installment we saw how to compute a common logarithm: simply dial in the argument on the C, D, CI or DI scale, and read the log on the L scale.
0 kommentar(er)
