also: tinyurl.com/griffenfly

Finding Logs and Powers of 10 with the L Scale

These instructions cover the use of:
  1. finding Finding a Log base 10
  2. Finding Powers of 10
Similar to looking up squares and roots with the B scale, the L scale lets you look up logs and powers of 10, depending on whether you read from C to L, or from L to C. And also like the B scale, the hard part isn't doing the lookup, it's setting the decimal place. For many problems you already have an idea of the decimal place, and the lookup is very fast. These rules are step by step instructions you can apply for when you have no idea what the decimal place is, or when you suspect you might be wrong.

Terminology

We will assume our slide rule has L on the slider and use C,CI, in these explanations. If your L scale is on the stock, use D,DI instead.

In this tutorial, we always assume the C,CI scales run from 1 to 10. We will refer to numeric positions on C,CI as decimal numbers between 1 and 10, such is 1.23CI. Numbers on the L scale will be between 0 and 1.

Remember that the CI scales run backwards, so 10CI is the left index, not the right.

We use the term mantissa to refer to the decimal part of a number in standard form, so the mantissa of 3.77 is 0.77. In some literature the scientific notation coefficient is referred to as a mantissa, but not in these tutorials, so don't confuse the two.

Finding base 10 Logarithms with L

To find y = log (base 10) of x using L and C

  • convert x to scientific notation.
  • read mantissa on L at xC
  • y = mantissa + x's exponent.
If x is < 1, that last step might be time consuming to do in your head ( ie -1 + .149932 ) Instead, you can use L and CI as follows:

To find y = log (base 10) of x, using L and CI

  • convert x to scientific notation.
  • read mantissa on L at xCI
  • y = -mantissa + x's exponent + 1
Either the L,C or the L,CI methods can be used to find any logs, but the addition step is made trivial if you use C for x >= 1, and CI for 0 < x < 1.

Examples

log 2
2 is on C, we can just read the result off L: 0.301
or apply the L,C method: 2 -> 2E0, read 0.301 on L at 2C, 0 + 0.301 = 0.301.
or apply the L,CI method: 2 -> 2E0, read 0.699 on L at 2CI, -0.699 + 0 + 1 = 0.301.
log 20, L,C method
20 -> 2E1
mantissa on L at 2C = 0.301
.301 + x's exponent = 0.301 + 1 = 1.301
log 0.2, using L,C method
0.2 -> 2E-1
mantissa on L at 2C = 0.301
0.301 - 1 = -0.699 (in your head. This is the hard bit )
log 0.2, using L,CI method
0.2 -> 2E-1
mantissa on L at 2CI = 0.699
-0.699 + (-1) + 1 = -0.699 (much easier)
log 0.02, using L,CI method
0.02 -> 2E-2
mantissa on L at 2CI = 0.699
-0.699 + (-2) + 1 = -1.699

Finding a power of 10 with L

To find y = 10 raised to x, using L and C

  • split x into an integer xi and a positive mantissa xm
  • read coefficient of y on C at xmL
  • exponent of y is xi
  • optionally convert back to standard form.

To find y = 10 raised to x, using L and CI

  • split x into an integer xi and a negative mantissa xm
  • take absolute value of xm
  • read coefficient of y on CI at xmL
  • exponent of y is xi-1
  • optionally convert back to standard form.
These rules are the inverse of the logarithm rules. Either can be used to take any power of ten, but the L,C method is simple for positive powers with lots of decimal places,, and the L,CI method is simple for negative powers with lots of decimal places.

You might think that you won't be called upon to find many non-integer powers of ten in your day to day work, but these methods can be combined with the methods for finding logs base 10 to find logs and powers of any base using the just L and C scales. This is useful if your slide rule doesn't have LL scales, or if the values are offscale on the LL's.

Examples

10^1.8 using L,C method
xi = 1, xm = 0.8
coefficient of answer is on C at 0.8L : 6.31
answer is 6.31E1 = 63.1
10^0.02 using L,C method
xi = 0, xm = 0.02
coefficient of answer is on C at 0.02L : 1.047
answer is 1.047E0 = 1.047.
10^-2.4 using L,C method
xi = -2, xm = -0.4, <-notice xm is negative! this is the kicker.
we take 1 from xi and add it to xm to make it positive.
xi = -3, xm = 0.6. (still the same number: -3+.6 = -2-.4 )
coefficient of answer is on C at 0.6L : 3.98
answer is 3.98-3 = 0.00398
10^1.8 using L,CI method
xi = 1, xm = 0.8
take one from xm and add it to xi: xi=2, xm = -0.2
coefficient of answer is on CI at 0.2L : 6.31
subtract 1 from xi : 2-1 = 1
answer is 6.31E1 = 63.1
10^-2.4 using L,CI method
xi = -2, xm = -0.4, |-0.4| = 0.4
coefficient of answer is on CI at 0.4L : 3.98
subtract 1 from xi: -2-1 = -3
answer is 3.98E-3