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For a long time I've eschewed bulky and inelegant calculators for the use of my trusty trig/log-log slide rule. For those unfamiliar, here is a simple slide rule simulator using Javascript.

To demonstrate, find the $LL_3$ scale, which is on the back of the virtual one. Let's say we want to solve $3^n$.

First, you would move the cursor (the red line) over where $3$ is on the $LL_3$ scale. Then, you would slide the middle slider until the $1$ on the $C$ scale is lined up to the cursor.

And voila, your slide rule is set up to find $3^n$ for any arbitrary $n$. For example, to find $3^2$, move the cursor to $2$ on the $C$ scale, and your answer is what the cursor is on on the $LL_3$ scale ($9$). Move your cursor to $3$ on $C$, and it should be lined up with $27$ on $LL_3$. To $4$ on C, it is on $81$ on $LL_3$.

You can even do this for non-integer exponents ($1.3,\cdots$ etc.)

You can also do this for exponents less than one, by using the $LL_2$ scale. For example, to do $3^{0.5}$, you would find $5$ on the $C$ scale, and look where the cursor is lined up at on the $LL_2$ scale (which is about $1.732$).

Anyways, I was wondering if anyone could explain to me how this all works? It works, but...why? What property of logarithms and exponents (and logarithms of logarithms?) allows this to work?

I already understand how the basics of the Slide Rule works ($\ln(m) + \ln(n) = \ln(mn)$), with only multiplication, but this exponentiation eludes me.

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+1 just for the javascript sliderule link :-) – Jason S Jul 24 '10 at 2:22
up vote 7 down vote accepted

If x = 3n, then log x = n log 3.

The C scale is logarithmic, which means if the reading is p, then the distance is proportional to   log p.

Similarly, in the LLx scale the distance is proportional to   log log p.

Thus, when you align 1 to "3" in LL3, you introduce an offset of (log log 3). Suppose you get a reading of n in the C scale, then the corresponding value in LL3 would be:

log log p = log log 3 + log n
  (LL3)      (offset)    (C)

eliminating one level of log gives

    log p = log 3 * n

eliminating one more level of log gives

        p = 3^n

LL2 is the same as LL3 except it covers a different range.

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In the second equation, is it log(3*n) or log(3)*n? Either way, can you explain in more detail how you got the third one? I'm not sure how it follows. Thank you for the help though! – Justin L. Jul 23 '10 at 20:40
It's the second one and the answer to your other question, first he moved n inside the log where it became 3^n, the he raised both sides to the base of the log. Look here for more rules – Jonathan Fischoff Jul 23 '10 at 20:52

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