
Finding Logs and Powers of 10 with the L Scale
These instructions cover the use of:
 finding Finding a Log base 10
 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.
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 > 2E1
 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 > 2E1
 mantissa on L at 2CI = 0.699
 0.699 + (1) + 1 = 0.699 (much easier)
 log 0.02, using L,CI method
 0.02 > 2E2
 mantissa on L at 2CI = 0.699
 0.699 + (2) + 1 = 1.699

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 xi1
 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 noninteger 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 3.983 = 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 : 21 = 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: 21 = 3
 answer is 3.98E3

