Tutorial: Exponentiation and Logarithms with the LL Scales

Exponentiation and Logarithms with the LL Scales

These instructions cover:

Exponentiation
Logarithms

Introduction

If you've read the tutorial on the L scale, you'll know that you can find any log or exponent with just L and C. The problem with the L scale method is it takes multiple steps and you have to figure out the decimal place. The LL scales allow finding logs and exponents, with the correct decimal place in an easy procedure akin to doing a multiplication.

The price of convenience is loss of generality. You can only work with exponents that are on the LL scales, so there are a lot of LL scales to compensate. You can get by with 3, as on the Griffenfly Darmstadt, or throw 8 or even 10 on there.

We provide 18 different LL scales, but this doesn't mean there's something new to learn about each one. They are like the cube root Q1,Q2,Q3 scales - all the same scale but split into multiple lengths so they fit. And part of the reason for the large number is there are two types: those which are in base e in relation to C, (LL..E) and those which are in base 10 (LL..X). What's more, half of them are just reciprocal scales of the other half, for extra convenience.

Typically you would choose to include a set of base e LL scales or a set of base 10 LL scales, but not both. It doesn't matter which ones you use, they do the same thing. The base is irrelevant for logs and powers. The primary difference is that the base 10 LL..X scales give greater range, and base e LL..E scales give greater accuracy.

Likewise you may choose to include a complement of the reciprocal LL0 scales or not. When finding fractional or negative exponents these will save you steps, but they aren't strictly necessary.

Nomenclature

The LL scale names are historical, and confusing to say the least. Luckily you don't really bother with the names when using the scales, just whether to move up one or down one.

base e
positive powers of e	negative powers of e (reciprocals)
LL0_E 1.001 .. 1.01	LL00_E 0.999 .. 0.99
LL1_E 1.01 .. 1.1	LL01_E 0.99 .. 0.90
LL2_E 1.1 .. 2.7	LL02_E 0.90 .. 0.37
LL3_E 2.7 .. 22000	LL03_E 0.37 .. 0.000045

base 10

positive powers of 10 negative powers of 10 (reciprocals)

LL0_X 1.00023 .. 1.0023 LL00_X 0.99977 .. 0.9977

LL1_X 1.0023 .. 1.023 LL01_X 0.9977 .. 0.977

LL2_X 1.023 .. 1.26 LL02_X 0.977 .. 0.79

LL3_X 1.26 .. 10 LL03_X 0.79 .. 0.1

LL4_X 10 .. 10 billion LL04_X 0.1 .. 1 ten billionth

Note that the _E and _X suffixes show up only in the designer panel and the auto-doc card. The scale name printed on the slide rule doesn't show them. Thus LL3_X and LL3_E both appear as LL3.

Design Considerations

When deciding which scales to include, there are a few things to keep in mind.

-Keep them in order.
-Don't mix the bases.
-make sure they are all on the stock, or all on the slider.
-When paring down the list, discard LL0 and LL00 first.: since there's a limit approximation:; that can be used for those values.; Thus for 1.0005^5, just multiply .0005x5 and add the 1 back in: 1.0025. (but don't try using this if the base isn't very close to 1!); (This reduction to multiplication in the limit is plainly obvious on the LL0_E scale.)
-If you can live with finding lots of reciprocals, discard the negative powers scales.

Instructions

If the instruction asks to set 1C at 2LL, you'll have to find the LL scale which contains 2.0. If you're using base 10 scales, this will be on LL3, and if you're using base e scales it will be on LL2. The particular scale names don't matter, and we'll avoid using them as much as possible. If the instruction requires you to "go up a scale", it means to find a value on the next highest numbered scale than the one you are currently on. For instance going up from LL1 would be LL2. Going up from LL02 would be LL03. Likewise for "going down a scale". If you can go up or down no further, you can't find the result using the LL scales.

When instructed to "switch to the reciprocal of the scale", it means to move to the similarly numbered LL scale running in the opposite direction. LL3 -> LL03, LL01 -> LL1, etc.

In these instructions, we'll mostly use the C scale, with LL scales in base e, on the stock. If the instruction calls for setting 1C at 4.0 LL, you could also slide 4.0 LL until it's over 1D. All other instructions are the same.

In all cases, the number is read directly off the LL scale, there is no further requirement to set the decimal place.

Exponentiation

Method 1: to find y = z ^ x with reciprocal LL scales on the stock:

Set 1C or 10C over z LL, whichever keeps (coefficient of)x C onscale
if x is not on C, go up n scales, where n is x's scientific notation exponent.
if you used the 10C index, go up one more.
if x is positive, read answer at x C on the LL scale you're now on.
if x is negative, read answer at x C on the reciprocal of the scale you're now on.

But not all slide rules have complimentary reciprocal LL scales. Look at the back face of the Griffenfly Darmstadt for an example. You can still calculate powers, if you note that a^b and a^(-b) are reciprocals, as are a^b and (1/a)^b. The following method works, but it adds more complexity to the calculations and reduces accuracy and speed.

Method 2: To find y = z ^ x with no reciprocal LL scales:

if z > 1 and x > 0 use Method 1.
if z > 1 and x < 0 use Method 1 with |x|, and take reciprocal of answer.
if z < 1 and x > 0 use Method 1 with 1/z, and take reciprocal of answer.
if z < 1 and x < 0 use Method 1 with 1/z and |x|.

( Note that by z < 1 we mean z is between 0 and 1. Negative numbers raised to fractional powers are not necessarily on the Real number line, for instance (-1)^(0.5) has two answers at i and -i. All you can do is use the absolute value of z and keep in mind that you're only getting the magnitude, not the direction nor the number of answers. )

Examples (mostly using Method 1)

3^2: (this one uses base e scales); find 3 on the LL stack. It's near left end on LL3.; set 1C at 3 LL3; 2 is found on C, so we stay on LL3; we used 1C, so we stay on LL3; read answer 9 on LL3 at 2C

3^(0.2): (using base e scales); find 3 on the LL stack. It's near left end on LL3.; set 1C at 3 LL3; 0.2 is 2E-1 so we "move up -1", ie move down 1 to LL2.; we used 1C, so we stay on LL2; read answer 1.246 on LL3 at 2C

2^10: (using base e scales); find 2 on the LL stack. It's near right end on LL2.; Set cursor there, read answer 1024 on LL3.

This last example is a shortcut. The values on one LL scale are the values on the previous scale raised to the power of 10. This is true regardless of whether base e or base 10 scales are used.

0.51 ^ 3.1 (using base e scales including reciprocal LL0 scales. ): find 0.51 on the LL stack. It's near the right end of LL02; set 10C on 0.51 LL02 (notice it runs right to left); 3.1 is on C, so we stay on LL02; we used 10C, so we move up a scale to LL03; read answer 0.124 on LL03.
1.07 ^ -31.9 (using base e scales including reciprocal LL0 scales. ): find 1.07 on the LL stack. (on LL1), set 10C over it to keep 3.19 C onscale.; -31.9 = -3.19E1, so we move up to LL2; we used 10 C so we move up again to LL3; x is negative, so we switch to the reciprocal scale, LL03.; read answer 0.1155 on LL03 at 3.19C

The above example only works because the base is so close to 1. We couldn't have found 10.7^-31.9. In general large bases and large exponents will be off the LL scales and you'll have to resort to the L scale method of exponentiation. After working with floating points on so many other slide rule techniques, you might be tempted to think you can just shift the factor of ten, but that's not the case with LL scales.

0.05 ^ 2.5 ( using base 10 scales, no reciprocal scales, on the slider ): We'll have to use method 2, because 0.05 < 1 and we have no reciprocal LL scales.; z < 1 and x > 0, so we use the third row in Method2; use D/DI to take reciprocal of 0.05 -> 20.; 20 is on LL4 near the left side.; LL4 is on the slider now, so instead of setting 1C at 20 LL4, we have to set 20 LL4 at 1D.; We used 1D, so we stay on LL4; 2.5 is on D, so we stay on LL4; read is 1789 on LL4, at 2.5 D; we're using third row of method 2, so we take reciprocal of this using D,DI: 5.590E-4, 0.0005590.

This last example coveres everything we didn't cover in the previous examples. Moving up to the slider isn't so hard, nor is using the base 10 scales. However taking two sets of reciprocals makes things tedious and introduces error because of the extra steps. ( Note: In the diagram, those extra little "reciprocal cursors" are just drawn in, they're not part of the slide rule. )

Finding Logs

Before we show the method for finding logs to any power, we should say that logs to base 10 or base e are available by direct lookup, from the LL scales to the C or D scale. However the result has to be modified by the exponent in the auto doc card. For instance, if you're using base e LL scales, to find Ln 0.8, find it on the LL02 scale, set the cursor there, and look up the answer, 2.2314 on C. Then read the auto doc card for LL02 ( hover the mouse over the right scale name ), and note the exponent of e. It says -x/10. So we apply that to 2.2314 and get -0.22314, the correct answer.

( This is the functionality that is extended by including the CFM or CF/M scales. )

Method 1: to find x = log(base z) y with reciprocal LL scales on the stock:

Set 1C or 10C over z LL, whichever keeps (coefficient of)y C onscale
Initial result x is on C, at y on LL.
if you used 10C, reduce answer's scientific notation exponent by 1.
subtract y's LL number from z's LL number. Increase exponent by this amount
if y's scale runs in a different direction to z's scale, negate x.

Method 2: to find x = log(base z) y with no reciprocal LL scales:

if z > 1 and y > 1 use Method 1.
if z > 1 and y < 1 use Method 1 with 1/y, negate the answer.
if z < 1 and y > 1 use Method 1 with 1/z, negate the answer.
if z < 1 and y < 1 use Method 1 with 1/y and 1/z.

Examples ( mostly using base 10 LL scales on the stock, using method 1 )

log(base5) 500: set 10C over the base: 5, found on LL3; set cursor at 500 - found on LL4; read 3.861 off of C; we used 10C, so divide by ten: 0.3681; LL4-LL3 = 1, so raise the exponent by 1: 3.681; Ll4 and LL4 both run left to right, so we leave the answer.
log(base 1,000,000) 0.9: set 10C over base 1M on LL4; set cursor at 0.9 on LL02; read 7.63 on C; we used 10C so divide by 10: 0.763; LL02 - LL4: 2-4 = -2, divide by 100: 0.00764; LL02 runs opposite to LL4, so negate the answer: -0.00763, -7.63E-3.
log( base 0.5 ) 0.7 using base e scales on the stock, no reciprocal scales:: both numbers are less than 1, so we can just convert to reciprocals and use method 1.; 1/0.5= 2, 1/0.7 = 1.4286 (use D,DI scales); find log( base 2 ) 1.4286:; set 2 on LL2 over 10 D; read 5.146 on D at 1.4286 LL2; we used 10D, so divide by 10: 0.5146; we didn't change scales or direction, so answer is correct.