# Why does the log-log range on my Slide Rule job?

For a long period of time I've shunned cumbersome and also inelegant calculators for making use of my dependable trig/log-log slide regulation. For those strange, below is a straightforward slide rule simulator using Javascript.

To show, locate the $LL_3$ range, which is on the back of the digital one. Allow's claim we intend to address $3^n$.

First, you would certainly relocate the arrow (the red line) over where $3$ gets on the $LL_3$ range. After that, you would certainly *glide* the center slider till the $1$ on the $C$ range is aligned to the arrow.

And also voila, your slide regulation is set up to locate $3^n$ for any kind of approximate $n$. As an example, to locate $3^2$, relocate the arrow to $2$ on the $C$ range, and also your solution is what the arrow gets on on the $LL_3$ range ($9$). Relocate your arrow to $3$ on $C$, and also it needs to be associated $27$ on $LL_3$. To $4$ on C, it gets on $81$ on $LL_3$.

You can also do this for non-integer backers ($1.3,\cdots$ and so on )

. You can additionally do this for **backers much less than one**, by utilizing the $LL_2$ range. As an example, to do $3^{0.5}$, you would certainly locate $5$ on the $C$ range, and also look where the arrow is aligned at on the $LL_2$ range (which has to do with $1.732$).

Anyways, I was asking yourself if any person could clarify to me just how this all jobs? It functions, yet ... why? What building of logarithms and also backers (and also logarithms of logarithms? ) permits this to function?

I currently recognize just how the essentials of the Slide Rule functions ($\ln(m) + \ln(n) = \ln(mn)$), with just reproduction, yet this exponentiation thwarts me.

If x = 3 ^{n }, after that log x = n log 3.

The C range is logarithmic, which suggests if the analysis is p, after that the range is symmetrical to log p.

Similarly, in the LLx range the range is symmetrical to log log p.

Thus, when you straighten 1 to "3" in LL3, you present an *countered * of (log log 3). Intend you get an analysis of *n * in the C range, after that the equivalent value in LL3 would certainly be:

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

removing one degree of log offers

```
log p = log 3 * n
```

removing another degree of log offers

```
p = 3^n
```

LL2 coincides as LL3 other than it covers a various array.

Related questions